Understanding Redox Reactions

SUGGESTED ANSWERS
ANSWER CHAPTER 1
Activity 1A (Page 5)
1. (a) 2CuO + C → 2Cu + CO2
(b) (i) Reduction reaction (ii) Oxidation reaction
(c) (i) Carbon (iii) Copper(II) oxide
(ii) Copper(II) oxide (iv) Carbon
Activity 1B (Page 6)
(a) (i) Ammonia (iii) Copper(II) oxide
(ii) Copper(II) oxide (iv) Ammonia
(b) - Ammonia is oxidised because loses hydrogen
- Copper(II) oxide is reduced because loses oxygen.
- Copper(II) oxide is an oxidising agent because causes ammonia to undergo oxidation reaction.
- Ammonia is a reducing agent because causes copper(II) oxide to undergo reduction reaction.
Laboratory Activity 1A (Page 7)

1. X : Fe2+ → Fe3+ + e-
Y : MnO4- + 8H+ + 5e- → Mn2+ + 4H2O
2. X : Oxidation
Y : Reductionn
3. 5Fe2+ + MnO4- + 8H+ → 5Fe3+ + Mn2+ + 4H2O
4. - Fe2+ ion is oxidised because loses electron.
- MnO4- ion is reduced because gains electron.
- Oxidising agent is MnO4- ion because MnO4- ion is the electron acceptor.
- Reducing agent is Fe2+ ion because Fe2+ ion is the electron donor.
5. Electron move from X electrode to Y electrode through connecting wire.
6. Positive terminal is Y electrode and negative terminal is X electrode.
7. Allow the transfer of ion and to complete the curcuit.
8.
G
Carbon electrode Carbon electrode
Chlorine water Potassium iodide

solution
Activity 1C (Page 11)
CuO + H2 → Cu + H2O
(a) Redox reaction occurs because oxidation number of copper decreases from +2 to 0 while
oxidation number of hydrogen increases from 0 to +1.
(b) (i) H2 is oxidised and CuO is reduced.
(ii) Oxidising agent is CuO and reducing agent is H2.
Mg + 2HCl → MgCl2 + H2
(a) Redox reaction occurs because oxidation number of magnesium increases from 0 to +2 while
oxidation number of hydrogen decreases from +1 to 0.
(b) (i) Mg is oxidised and HCl is reduced.
(ii) Oxidising agent is HCl and reducing agent is Mg.
Activity 1D (Page 13)
1. +7 3. -1 5. +1 7. +4
2. +2 4. +1 6. +3 8. 0
Activity 1E (Page 14)

1. (a) Copper(I) oxide (d) Mercury(II) chloride
(b) Copper(II) oxide (e) Potassium chromate(VI)
(c) Mercury(I) chloride
2. (a) V2O5 (b) Na2O (c) PbCO3
3. Fe2O3 : Iron(III) oxide
Al2O3 : Aluminium oxide
- Nomenclature of compounds Fe2O3 has Roman numeral while Al2O3 does not has Roman
numeral.
- Iron has more then one oxidation number and Roman numeral at iron(III) oxide shows the
oxidation number of iron is +3 while aluminium has only one oxidation number so does not
need Roman numeral.
Laboratory Activity 1B (Page 15)

Result:
Table to record observation in Sections A and B.
Mixture Observation
Iron(II) sulphate solution + Bromine water
Iron(III) chloride solution + Zinc powder
1. Section A :.
(a) Fe2+ ion is oxidised because oxidation number of iron increases from +2 to +3 while
bromine water is reduced because oxidation number of bromine decreases from 0 to -1.
(b) Fe2+ ion is oxidised because Fe2+ ion loses electron to formed Fe3+ ion and bromine water is
reduced because bromine molecule, Br2 gains electron to formed bromide ion, Br-.
Section B :
(a) Zinc is oxidised because oxidation number of zinc increases from 0 to +2 while Fe3+ ion is
reduced because oxidation number of iron decreases from +3 to +2.
(b) Zinc is oxidised because zink atom, Zn loses electron to formed Zn2+ ion and Fe3+ ion is
reduced because Fe3+ gains electron to formed Fe2+ ion.
2. - Bromine water is an oxidising agent because bromine is the electron acceptor.
- Zinc is a reducing agent because zinc is the electron donor.
3. Iron(II) sulphate solution is easily oxidised to iron(III) sulphate.
4. Sodium hydroxide solution.
5. Section A : Chlorine water
Section B : Magnesium powder
Laboratory Activity 1C (Page 18)

Prosedure:
1. Clean magnesium ribbon, lead plate and copper plate using sand paper.
2. Put magnesium ribbon into two difference test tubes.
3. Pour 0.5 mol dm-3 lead(II) nitrate solution into the first test tube and 0.5 mol dm-3 copper(II)
nitrate solution into the second test tube. The salt solution is poured until the magnesium ribbon is
immersed.
4. Repeat step 2 using lead plate and copper plate.
5. Pour 0.5 mol dm-3 magnesium(II) nitrate solution into the third test tube and 0.5 mol dm-3
copper(II) nitrate solution into the fourth test tube. The salt solution is poured until the lead plate
is immersed.
6. Pour 0.5 mol dm-3 magnesium(II) nitrate solution into the fifth test tube and 0.5 mol dm-3 lead(II)
nitrate solution into the sixth test tube. The salt solution is poured until the copper plate is
immersed.
7. Put all the test tubes on the test tube rack.
8. Record the observation.
Result:
Table to record observation and inference of metal displacement reaction.
Metal Salt solution Observation Inference
Lead(II) nitrate
Magnesium
Copper(II) nitrate
Magnesium nitrate
Lead
Copper(II) nitrate
Magnesium nitrate
Copper
Lead(II) nitrate
1. Experiment: Mg + Pb(NO3)2
(a) Oxidation half equation : Mg → Mg2+ + 2e-
Reduction half equation : Pb2+ + 2e- → Pb
Overall ionic equation : Mg + Pb2+ → Mg2+ + Pb
(b) (i) Magnesium (iii) Lead(II) nitrate
(ii) Lead(II) nitrate (iv) Magnesium
(c) (i) Magnesium atom loses electron to formed Mg2+ ion.
(ii) Pb2+ ion gains electron to formed lead, Pb atom.
(iii) Pb2+ ion is the electron acceptor
(iv) Mg is the electron donor
Experiment: Mg + Cu(NO3)2
(a) Oxidation half equation: Mg → Mg2+ + 2e-
Reduction half equation: Cu2+ + 2e- → Cu
Overall ionic equation : Mg + Cu2+ → Mg2+ + Cu
(b) (i) Magnesium (iii) Copper(II) nitrate
(ii) Copper(II) nitrate (iv) Magnesium
(c) (i) Magnesium atom loses electron to formed Mg2+ ion.
(ii) Cu2+ ion gains electron to formed copper, Cu atom.
(iii) Cu2+ ion is the electron acceptor
(iv) Mg is the electron donor
Experiment: Pb + Cu(NO3)2
(a) Oxidation half equation: Pb → Pb2+ + 2e-
Reduction half equation: Cu2+ + 2e- → Cu
Overall ionic equation : Pb + Cu2+ → Pb2+ + Cu
(b) (i) Lead (iii) Copper(II) nitrate
(ii) Copper(II) nitrate (iv) Lead
(c) (i) Pb atom loses electron to formed Pb2+ ion.
(ii) Cu2+ ion gains electron to formed copper, Cu atom.
(iii) Ion Cu2+ ion is the electron acceptor
(iv) Pb is the electron donor
Laboratory Activity 1D (Page 20)

Result:
Table to record observation
Observation
Halide solution Halogen
Aqueous layer 1,1,1-trichloroethane layer
Bromine water
Potassium chloride
Iodine solution
Chlorine water
Potassium bromide
Iodine solution
Chlorine water
Potassium iodide
Bromine water
1. To verify the present of halogen.
2. (a) No
(b) Chlorine water.
(c) Chlorine water and bromine water.
3. KBr dan Cl2
(a) Oxidation half equation : 2Br- → Br2 + 2e-
Reduction half equation : Cl2 + 2e- → 2Cl-
(b) Overall ionic equation : 2Br- + Cl2 → Br2 + 2Cl-
(c) - Br- ion is oxidised because Br- ion loses electron to formed bromine, Br2 molecule while
chlorine is reduced because chlorine, Cl2 molecule gains electron to formed Cl- ion.
- Chlorine water is an oxidising agent because chlorine is the electron acceptor while
potassium bromide is a reducing agent because Br- ion is the electron donor.
KI dan Cl2
(a) Oxidation half equation : 2I- → I2 + 2e-
(b) Overall ionic equation : 2I- + Cl2 → I2 + 2Cl-
(c) - I- ion is oxidised because. I- ion loses electron to formed iodine, I2 molecule while chlorine
is reduced because chlorine, Cl2 molecule gains electron to formed chloride, Cl- ion.
- Chlorine water is an oxidising agent because chlorine is the electron acceptor while
potassium iodide is a reducing agent because I- ion is the electron donor.
KI dan Br2
Reduction half equation : Br2 + 2e- → 2Br-
(b) Overall ionic equation : 2I- + Br2 → I2 + 2Br-
(c) - I- ion is oxidised because I- ion loses electron to formed iodine, I2 molecule while bromine
is reduced because bromine, Br2 molecule gains electron to formed bromide, Br- ion.
- Bromine water is an oxidising agent because bromine is the electron acceptor while
potassium iodide is a reducing agent because I- ion is the electron donor.
4. (a) I2, Br2, Cl2
(b) The strength of halogen as an oxidising agent decreases when going down Group 17.
(c) Cl-, Br-, I-
Self Assess 1.1 (Page 21)
1. A chemical reaction where oxidation and reduction process occur simultaneously.
2. Reaction I
(a) Oxidation half equation : Cu → Cu2+ + 2e-
Reduction half equation : Ag+ + e- → Ag
(b) - Copper is oxidised because copper, Cu atom loses electron to formed Cu2+ ion while Ag+
ion is reduced because Ag+ gains electron to formed silver, Ag atom.
- Ag+ ion is an oxidising agent because Ag+ ion is the electron acceptor while copper is a
reducing agent because copper is the electron donor.
Reaction II
(a) Oxidation half equation : Pb → Pb2+ + 2e-
Reduction half equation : O2 + 4e- → 2O2-
(b) - Lead is oxidised because lead, Pb atom loses electron to formed Pb2+ ion while oxygen is
reduced because oxygen, O2 molecule gains electron to formed oxide, O2- ion.
- Oxygen is an oxidising agent because oxygen is the electron acceptor while lead is a
reducing agent because lead is the electron donor.
Reaction III
(a) Oxidation half equation : Al → Al3+ + 3e-
(b) - Aluminium is oxidised because aluminium, Al atom loses electron to formed Al3+ ion
while chlorine is reduced because chlorine, Cl2 molecule gains electron to formed
chloride, Cl- ion.
- Chlorine is an oxidising agent because chlorine is the electron acceptor while
aluminium is a reducing agent because aluminium is the electron donor.
Reaction IV
Reduction half equation : Br2 + 2e- → 2Br-
(b) - I ion is oxidised because I ion loses electron to formed iodine, I2 molecule while bromine
- -
is reduced because bromine, Br2 molecule gains electron to formed bromide, Br- ion.
- Bromine water is an oxidising agent because bromine is the electron acceptor while iodide
ion is a reducing agent because I- ion is the electron donor.
Activity 1F (Page 26)

2. (a) - The standard electrode potential value, Eo is arranged from the most negative to the most
positive
Cr2O72-(aq) + 14H+(aq) + 6e- ⇌ 2Cr3+(aq) + 7H2O(l) Eo = +1.33 V
Cl2(g) + 2e- ⇌ 2Cl- (aq) Eo = +1.36 V
- The Eo value of Cr2O72- is less positive therefore Cr2O72- ion on the left side is a weaker
oxidising agent. Cr2O72- ion is difficult to gains electron and reduction reaction does not
occur.
- The Eo value of Cl- is more positive therefore Cl- ion on the right side is a weaker reducing
agent. Cl- ion is difficult to loses electron and oxidation reaction does not occur. Therefore,
the reaction between Cr2O72- and Cl- does not occur.
(b) - The standard electrode potential value, Eo is arranged from the most negative to the most
positive.
Br2(l) + 2e- ⇌ 2Br- (aq) Eo = +1.07
H2O2(aq) + 2H (aq) + 2e ⇌ 2H2O(l)
+ -
Eo = +1.77
- The Eo value of H2O2 is more positive therefore H2O2 on the left side is a stronger
oxidising agent. H2O2 gains electron easily and reduction reaction occurs.
- The Eo value of Br- is less positive therefore Br- ion on the right side is a stronger reducing
agent. Br- ion loses electron easily and oxidation reaction occurs. Therefore, the reaction
between H2O2 and Br- occurs.

1. (a) Arranged the standard electrode potential value, Eo from the most negative to the most
positive.
Mg2+(aq) + 2e- ⇌ Mg(p) Eo = -2.38
Zn (aq) + 2e ⇌ Zn(p)
2+ -
Eo = -0.76
Cu2+(aq) + 2e- ⇌ Cu(p) Eo = +0.34
Ag+(aq) + e- ⇌ Ag(p) Eo = +0.80
Oxidising agent : Mg , Zn , Cu , Ag
2+ 2+ 2+ +
Reducing agent : Ag, Cu, Zn, Mg

(b) (i) Reaction occur
- Mg is a stronger reducing agent. Mg atom loses electron easily and oxidation reaction
occurs.
- Cu2+ ion is a stronger oxidising agent. Cu2+ ion gains electron easily and reduction
reaction occurs.
- Redox reaction occurs between strong reducing agent and strong oxidising agent.
(ii) Reaction occur
- Mg is a stronger reducing agent. Mg atom loses electron easily and oxidation reaction
occurs.
- Zn2+ ion is a stronger oxidising agent. Zn2+ ion gains electron easily and reduction
reaction occurs.
- Redox reaction occurs between strong reducing agent and strong oxidising agent.
(iii) Reaction does not occur
- Cu is a weaker reducing agent. Cu atom is difficult to loses electron and oxidation
reaction does not occur.
- Zn2+ ion is a weaker oxidising agent. Zn2+ ion is difficult to gains electron and
reduction reaction does not occur.
- Redox reaction does not occur between weak reducing agent and weak oxidising
agent.
Experiment 1A (Page 29)

Sample answer: Simple chemical cell
Material: Magnesium ribbon, iron nail, zinc strip, lead strip, copper strip and copper(II) nitrate
solution.
Apparatus: Beaker
Prosedure:
1. Clean magnesium ribbon, iron nail, zinc strip, lead strip and copper strip using sand paper.
2. Pour 1.0 mol dm-3 copper(II) nitrate solution into a beaker until half full.
3. Connect magnesium ribbon and copper strip to a voltmeter using connecting wire.
4. Dip magnesium ribbon and copper strip into copper(II) nitrate solution to complete the circuit.
5. Record voltmeter reading, metal at the negative terminal and metal at the positive terminal.
6. Repat steps 3 to 5 using iron nail, zinc strip and lead strip to replace magnesium ribbon.
Magnesium ribbon Copper strip
Copper(II) nitrate
solution
OR
Sample answer: Chemical cell with two half cell.
Material: Magnesium ribbon, iron nail, zinc strip, lead strip, copper strip, magnesium nitrate solution,
iron(II) nitrate solution, zinc nitrate solution, lead(II) nitrate solution and copper(II) nitrate
solution.
Apparatus: Beaker and porous pot
Prosedure:
1. Rub magnesium ribbon, iron nail, zinc strip, lead strip and copper strip using sand paper.
2. Pour 1.0 mol dm-3 magnesium nitrate solution into a porous pot and 1.0 mol dm-3 copper(II)
nitrate solution into a beaker until half full.
3. Place the porous pot into the beaker.
4. Connect magnesium ribbon and copper strip to a voltmeter using connecting wire.
5. Dip magnesium ribbon into magnesium nitrate solution and copper strip into copper(II) nitrate
solution to complete the circuit.
6. Record voltmeter reading, metal at the negative terminal and metal at the positive terminal.
7. Repat steps 2 to 6 using iron(II) nitrate solution, zinc nitrate solution and lead(II) nitrate solution
to replace magnesium nitrate solution in the porous pot and using iron nail, zinc strip and lead
strip to replace magnesium ribbon.
Discussion:
1. - Pair of Mg/Cu
Oxidation half equation : Mg → Mg2+ + 2e-
Reduction half equation : Cu2+ + 2e- → Cu
Overall ionic equation : Mg + Cu2+ → Mg2+ + Cu
- Pair of Fe/Cu
Oxidation half equation : Fe → Fe2+ + 2e-
Overall ionic equation : Fe + Cu2+ → Fe 2+ + Cu
- Pair of Zn/Cu
Oxidation half equation : Zn → Zn 2+ + 2e-
Overall ionic equation : Zn + Cu2+ → Zn 2+ + Cu
- Pair of Pb/Cu
Oxidation half equation : Pb → Pb 2+ + 2e-
Overall ionic equation : Pb + Cu2+ → Pb 2+ + Cu
2. - Pair of Mg/Cu
Mg(s) | Mg2+(aq, 1 mol dm-3) || Cu2+(aq, 1 mol dm-3) | Cu(s)
- Pair of Fe/Cu
Fe(s) | Fe2+(aq, 1 mol dm-3) || Cu2+(aq, 1 mol dm-3) | Cu(s)
- Pair of Zn/Cu
Zn(s) | Zn2+(aq, 1 mol dm-3) || Cu2+(aq, 1 mol dm-3) | Cu(s)
- Pair of Pb/Cu
Pb(s) | Pb2+(aq, 1 mol dm-3) || Cu2+(aq, 1 mol dm-3) | Cu(s)
3. - Pair of Mg/Cu
E0cell = (+0.34) – (- 2.38) = + 2.72 V
- Pair of Fe/Cu
E0cell = (+0.34) – (- 0.44) = + 0.78 V
- Pair of Zn/Cu
E0cell = (+0.34) – (- 0.76) = + 1.10 V
- Pair of Pb/Cu
E0cell = (+0.34) – (- 0.13) = + 0.47 V
4. The greater the difference of E0 value for the pair of metals, the greater the
voltage of the cell.
Activity 1G (Page 30)

1. (a) (i) Negative terminal: Iron
Positive terminal : Silver
(ii) Fe(s) | Fe2+(aq) || Ag+(aq) | Ag(s)
(iii) Oxidation half equation : Fe → Fe2+ + 2e-
Reduction half equation : Ag+ + e- → Ag
Overall ionic equation : Fe + 2Ag+ → Fe 2+ + 2Ag
(iv) E0cell = (+0.80) – (- 0.44) = + 1.24 V

1. (a) Mg(s) | Mg2+(aq) || Sn2+(aq) | Sn(s)
(b) Pt(s) | Cl-(aq), Cl2(aq) || MnO4-(aq), Mn2+(aq) | Pt(s)
2. (a) E0cell = (+0.15) – (- 0.25) = + 0.40 V
(b) E0cell = (+0.80) – (+0.54) = + 0.26 V
(c) E0cell = (+0.80) – (+0.77) = + 0.03 V
(d) E0cell = (+1.36) – (+ 1.07) = + 0.29 V
Carbon electrode
Copper(II)
chloride
solution
Activity 1H (Page 33)
Apparatus set-up Observation Inference Conclusion

Bulb light up Presence of free Copper(II) chloride,
moving ion CuCl2 is electrolyte
Carbon electrode
Copper(II)
chloride
solution
Bulb light up Presence of free Copper(II) chloride,
moving ion CuCl2 is electrolyte
Carbon electrode
Crucible
Molten
copperII)
chloride
Bulb does not light up No free moving ion Glucose, C6H12O6 is

non-electrolyte
Carbon electrode
Glucose
solution
Bulb does not light up No free moving ion Glocose, C6H12O6 is
non-electrolyte
Carbon electrode
Crucible
Molten glucose
Bulb light up Presence of free Asid oksalik, C2H2O4

moving ion is electrolyte
Carbon electrode
Oxalic acid
Bulb light up Presence of free Ammonia , NH3 is

moving ion electrolyte
Carbon electrode
Ammonia
aqueous
Bulb does not light up No free moving ion Hexane, C6H14 is non-
electrolyte
Carbon electrode
Hexane
Bulb does not light up No free moving ion Ethanol, C2H5OH is
non-electrolyte
Carbon electrode
Ethanol
Laboratory Activity 1E (Page 34)

1. Lead(II) ion moves to cathode and bromide ion moves to anode.
2. (a) Pb2+ + 2e- → Pb
(b) 2Br- → Br2 + 2e-
3. Cathode : Lead
Anode : Bromine
4. Cathode : Lead(II) ion gains two electrons to form lead atom.
Anode : Bromide ion loses one electron to form bromine atom. Two bromine atoms combine
to form bromine molecule. (OR two bromide ions lose two electrons to form bromine
molecule)
5. Pb2+ + 2Br- → Pb + Br2
Activity 1I (Page 37)

1. (a)
Colourless gas released
Anode
Two oxide ions lose four electrons to
Molten form oxygen molecule
zinc oxide,
ZnO Grey solid deposited
Cathode
Zinc ion gains two electrons to form
zinc atom
(b) Greenish yellow gas released
Anode
Two chloride ions lose two electrons to
Molten
form chlorine molecule
Magnesium
chloride,
MgCl2 Grey solid deposited
Cathode
Magnesium ion gains two electrons to
form zinc atom
Laboratory Activity 1F (Page 38)

1. (a) Cu2+ ion, SO42- ion, H+ ion, OH- ion
(b) H+ ion, SO42- ion, OH- ion
2. Copper(II) sulphate solution
Cathode Anode
(a) Cu ion and H ion
2+ +
SO42- ion dan ion OH-
(b) Cu2+ ion OH ion
-
E0 value of Cu2+ ion is more positive E0 value of OH- ion is less positive compare
compare to E0 of H+ ion to E0 of SO42- ion
(c) Copper Oxygen
(d) Cu2+ + 2e- → Cu 4OH- → O2 + 2H2O + 4e-
(e) Cu2+ ion gains two electrons to form copper Four OH− ions lose four electrons to form
atom oxygen molecule and water
(f) 2Cu2+ + 4OH- → 2Cu + O2 + 2H2O
Sulphuric acid
Cathode Anode
(a) H ion
+
SO42- ion dan ion OH-
(b) H+ ion OH ion
-
Only H+ ion presents at cathode E0 value of OH- ion is less positive compare
to E0 of SO42- ion
(c) Hydrogen Oxygen
(d) 2H+ + 2e- → H2 4OH- → O2 + 2H2O + 4e-
(e) Two H+ ions gain two electrons to form Four OH− ions lose four electrons to form
hydrogen molecule oxygen molecule and water
(f) 4H+ + 4OH- → 2H2 + O2 + 2H2O
Experiment 1B (Page 41)

1. Hydrochloric acid, HCl 1.0 mol dm−3
(a) Chlorine.
Chloride ion is discharge because the concentration of Cl- ion is higher than OH- ion.
Two chloride ions lose two electrons to form chlorine molecule.
(b) 2Cl- → Cl2 + 2e-
Hydrochloric acid, HCl 0.0001 mol dm−3

(a) Oxygen
Hydroxide ion is discharge because the E0 value of OH- ion is less positive compare to E0
value of Cl- ion
Four OH− ions lose four electrons to form oxygen molecule and water.
(b) 4OH- → O2 + 2H2O + 4e-
2. Process that produced gas bubbles at anode when carbon electrodes connected to batteries are
dipped into hydrochloric acid.
3. Colourless gas released. Hydroxide ion is discharge because the E0 value of OH- ion is less
positive compare to E0 value of Cl- ion. Four OH− ions lose four electrons to form oxygen
molecule and water.
Experiment 1C (Page 42)
Hypothesis:
When carbon electrode is used, colourless gas bubbles released at anode while copper electrode is
used, anode copper thinner.
Variables:
a. Manipulated variable: carbon electrode and copper electrode
b. Responding variable: Observation at anode
c. Fixed variable: Copper(II) sulphate solution
Prosedure:
1. Pour 0.5 mol dm−3 of copper(II) sulphate solution, CuSO4 into a beaker until half full.
2. Connect carbon elektrodes to suis and batteries using connecting wire.
3. Dipped carbon elektrodes into copper(II) sulphate solution to complete the curcuit.
4. Observe and record changes occur at anode and cathode in a table.
5. Repeat steps 1 to 4 using copper electrode to replace carbon elektrode.
Result:
Observation
Type of electrode
Cathode Anod
Carbon
Copper
1. (a) Cathode : Copper
Anode : Oxygen
(b) Cathode : Copper
Anode : Copper(II) ion.
2. (a) Anode : 4OH- → O2 + 2H2O + 4e-
(b) Anode : Cu → Cu2+ + 2e-
3. (a) - Blue colour of copper(II) sulphate solution become paler.
- Concentration of Cu2+ ion decreases.
- Cu2+ ion is discharge to form copper atom at cathode.
(b) - Blue colour of copper(II) sulphate solution remain unchange.
- Concentration of Cu2+ ion remain unchange
- Rate of Cu2+ ion discharge to form copper atom at cathode is the same as the rate of copper
atom ionise to form Cu2+ ion at anode.
Activity 1J (Page 43)

2.
(a) Product
X : Chlorine
Y : Hydrogen
(b) Equation
X : 2Cl- → Cl2 + 2e-
Y : 2H+ + 2e- → H2
(c) Product if concentration of electrolyte 0.0001 mol dm−3

X : Oxygen
Y : Hydrogen
(b) (a) Product
J : Oxygen
K : Silver
(b) Equation
J : 4OH- → O2 + 2H2O + 4e-
K: Ag+ + e- → Ag
(c) Product if concentration of electrolyte 0.0001 mol dm−3

J : Oxygen
K: Silver
Activity 1K (Page 44)

1. - Electrolytic cell:
Anode(+) : Cu → Cu2+ + 2e-
Cathode(-) : Cu2+ + 2e- → Cu
- Sel kimia:
Anode(-) : Zn → Zn2+ + 2e-
Cathode(+) : Cu2+ + 2e- → Cu
- Blue colour of copper(II) sulphate solution remain unchange in electrolytic cell but blue colour
of copper(II) sulphate solution become paler in chemical cell.
- Concentration of Cu2+ ion remain unchange in electrolytic cell but concentration of Cu2+ ion
decreases in chemical cell.
- In electrolytic cell, rate of Cu2+ ion discharge at cathode is the same as the rate of copper
atom ionise at anode but in chemical cell, Cu2+ ion is discharge to form copper atom at
cathode.
Laboratory Activity 1G (Page 45)

1. (a) Yes.
(b) Oxidation reaction occurs when copper atom loses electron to form Cu2+ ion at anod
(kuprum) and reduction reaction occurs at cathode (iron spoon) when Cu2+ ion gains electron
to form copper atom.
2. Blue colour of copper(II) sulphate solution remain unchange because concentration of Cu2+ ion
remain unchange.
Rate of Cu2+ ion discharge at cathode is the same as the rate of copper atom ionise at anode.
3. - Rotate iron spoon during the electroplating process to get uniform layer of the plating
metal.
- Use small current during the electroplating process to get strong coating layer on the iron
spoon.
4.
Nikel Iron ring
Nikel nitrate solution

Laboratory Activity 1H (Page 46)
1. Yes. Oxidation reaction occurs at impure copper plate (anode) when copper atom
loses electron to form Cu2+ ion and reduction reaction occurs at pure copper plate (cathode) when
Cu2+ ion gains electron to form copper atom.
2. Blue colour of copper(II) nitrate solution remain unchange because concentration of Cu2+ ion
remain unchange.
Rate of Cu2+ ion discharge at cathode is the same as the rate of copper atom ionise at anode.
3. Anode is impure copper that contain impurities. When the impure copper ionises and dissolve,
the impurities form under the anode. Cathode is pure copper that does not has impurities.
4. Purification of copper metal is carried out through electrolysis process.

1. (a) Chloride ion and hydroxide ion
(b) Electrolytic cell I
- Oxygen gas.
- Hydroxide ion is discharge because the E0 value of OH- ion is less positive compare to E0
value of Cl- ion
- Four OH− ions lose four electrons to form oxygen molecule and water.
Electrolytic cell II
- Chlorine gas
- Cl- ion is discharge because the concentration of Cl- ion is higher than OH- ion.
- Two chloride ions lose two electrons to form chlorine molecule.
(c) (i) Cu → Cu2+ + 2e-
(ii) - Copper is active electrode.
- Copper atom ionises to form Cu2+ ion at anode.
(d) Oxidation reaction.
(e) Greenish yellow gas released
2. (a) Iron key looks more attractive / resistant to corrosion.
(b) (i) Silver nitrate solution.
(ii) Connect iron key to negative terminal of batteries and connect silver plate to positive
terminal of batteries.
3. (a)
Impure nikel Pure nikel
Nikel chloride solution
(b) Anode : Impure nikel metal becomes thinner.

Katod: Pure nikel metal becomes thicker
(c) Ni → Ni2+ + 2e-
Activity 1L (Page 49)
2. (a) During the electrolysis of molten aluminium oxide, Al3+ ion moves to cathode electrode
while O2- ion moves to anode. Al3+ ion is discharge to form aluminium while O2- ion is
discharge to form oxygen gas.
Half equation at cathode : Al3+ + 3e- → Al
Half equation at anode : 2O2- → O2 + 4e-
Ionic equation : 4Al3+ + 6O2- → 4Al + 3O2
(b) Electrolysis of molten aluminium oxide is carried out at high temperature so the oxygen gas
produce at anode reacts with carbon electrode to form carbon dioxide gas. Therefore, carbon
anode become thinner and should be replaced from time to time.
Chemical equation for the reaction involve:
2Al2O3 → 4Al + 3O2
O2 + C → CO2
(c) - Yes. Oxidation reaction at anode and reduction reaction at cathode occur simultaneously.
- At anode, O2- ion undergo oxidation reaction because O2- ion loses electron to form oxygen
molecule.
- At cathode, Al3+ ion undergo reduction reaction because Al3+ ion gains electron to form
aluminium atom.

1. Silver and aurum.
2. - Aluminium is more reactive than carbon in the reactivity series of metal.
- Reaction between aluminium oxide and carbon does not occurs.
- Carbon cannot reduced aluminium oxide.
3. (a) 2Fe2O3 + 3C → 4Fe + 3CO2
(b) - At high temperature, calcium carbonate, CaCO3 decomposes to form calcium oxide, CaO
(quicklime) and carbon dioxide, CO2.
- Calcium oxide, CaO reacts with impurities in iron ore such as silicone(IV)
oxide, SiO2 to form slag or calcium silicate, CaSiO3.
- This reaction is important to separate the impurities from molten iron.
- The difference in density causes slag to be at the top layer of the molten iron which
makes the separation process easier.
4. - Zinc (or Magnesium / Aluminium / Iron / Tin / Lead)
- Zinc is more reactive than copper.
- Zinc can reduce copper(II) oxide to copper.
Experiment 1D (Page 52)

Result:
Metal Observation Inference
Copper
Iron
1. To detect the present of Cu2+ ion and Fe2+ ion.

2. Copper: Cu → Cu2+ + 2e-
Iron : Fe → Fe2+ + 2e-
Fe2+ → Fe3+ + e-
3. Metal corossion occurs when metal atom loses electron to form metal ion.
Experiment 1E (Page 54)
Result:
Metal Observation
Fe/Mg
Fe/Sn
Fe/Zn
Fe/Cu
Fe
1. Potassium hexacyoanoferrate(III) solution is used to verify the present of Fe2+ ion and
phenolphthalein detects the present of OH- ion.
2. Blue spot and pink spot can be seen clearly and the colour does not mix.
3. The higher the intensity blue colour the higher the rate of rusting of iron.
4. As control.
5.
Metal prevent rusting of iron Metal accelerate rusting of iron
Magnesium Tin
Zinc Copper
6. Metal more electropositive than iron prevent rusting of iron while metal less electropositive than
iron accelerate the rate of rusting of iron.
7. Hypothesis accepted. When a more electropositive metal incontact with iron nail, iron nail does
not rust while metal less electropositive incontact with iron nail, iron nail rust.
Activity 1M (Page 57)

1. (a) Rusting of iron is a metal corrosion that occurs on iron only while metal corrosion occurs to
all metals.
(b) - Iron undergo oxidation reaction when iron atom loses electron to form Fe2+.
Fe → Fe2+ + 2e-
- Oxygen undergo reduction reaction when oxygen gains electron to form OH- ion.
O2 + 2H2O + 4e- → 4OH-
- Fe2+ ion reacts with OH− ion to form iron(II) hydroxide, Fe(OH)2.
Fe2+ + 2OH- → Fe(OH)2
- Fe(OH)2 undergoes continuous oxidation with oxygen to form hydrated iron(III) oxide
Fe2O3. xH2O , rust.
Fe(OH)2 → Fe2O3. xH2O
(c) - Metal more electropositive than iron has higher tendency to loses electron.
- Metal more electropositive than iron corrode and rusting of iron is prevented.
- Metal more electropositive than iron becomes a sacrificial metal.
- Examples: Bridge pillars, ship hulls and underground pipes.

1. (a) P : Copper
Q : Zinc
(b) Experiment I
- Redox reaction.
- Iron undergoes oxidation reaction when iron atom ferum loses electron to form Fe2+ ion.
Fe → Fe2+ + 2e-
- Blue spot in the test tube shows the present of Fe2+ ion.
- Oxygen undergoes reduction reaction when oxygen gains electron to form OH- ion.
O2 + 2H2O + 4e- → 4OH-
Experiment II
- Redox reaction.
- Zinc undergoes oxidation reaction when zinc atom loses electron to form Zn2+ ion.
Zn → Zn2+ + 2e-
- Oxygen undergoes reduction reaction when oxygen gains electron to form OH- ion.
O2 + 2H2O + 4e- → 4OH-
- Pink spot in the test tube shows the present of OH- ion.
- No blue spot in the test tube indicate no Fe2+ ion.
(c) Cu, Fe, Zn
2. - Tin forms a layer of oxide protection.
- Tin oxide layer protect steel from incontact with oxygen and water. Therefore steel does not
oxidised and rusting does not occurs.
Achievement Test (Page 59)

1. Reaction I is a redox reaction.
- Oxidation number of iron before reaction is +2 and after reaction is 0.
- Oxidation number of magnesium before reaction is 0 and after reaction is +2.
- Magnesium undergoes oxidation reaction because oxidation number of magnesium
increases from 0 to +2.
- Iron(II) sulphate undergoes reduction reaction because oxidation number of iron decreases from
+2 to 0.
Reaction II is not a redox reaction.
- Oxidation number all elements silver, nitrogen, oxygen, sodium and chlorine remain unchange
before and after the reaction.
2. (a) Oxidation number of iodine decreases from 0 to -1.
Oxidation number of sulphur increases from +4 to +6.
(b) Oxidising agent : Iodine.
Reducing agent : Sulphur dioxide.
(c) I2 + 2e- → 2I-
3. (a) Copper(II) nitrate solution.
Copper
(b) Oxidation half equation : Pb → Pb2+ + 2e-
(c) Pb + Cu2+ → Pb2+ + Cu
(d) +2 to 0
4. (a) Bromine
(b) Iodine
(c) Oxidation half equation: 2I- → I2 + 2e-
Reduction half equation: Br2 + 2e- → 2Br-
Ionic equation : 2I- + Br2 → I2 + 2Br-
5. Oxidising agent : R , Q , P2+
2+ 3+
Reducing agent : P, Q, R
6. (a) Negative terminal : Tin electrode.
Positive terminal : Platinum electrode.
(b) Sn(s) | Sn2+(aq, 1 mol dm-3) || Cl2(aq) , Cl-(aq, 1 mol dm-3) | Pt(s)
(c) Oxidation reaction : Sn → Sn2+ + 2e-
Reduction reaction: Cl2 + 2e- → 2Cl-
Ionic equation : Sn + Cl2 → Sn2+ + 2Cl-
(d) E0cell = (+1.36) – (- 0.14) = + 1.50 V
7. (a) Oxygen gas. Hydroxide ion is discharge because the E0 value of OH- ion is less positive
compare to E0 value of NO3- ion.
Four OH− ions lose four electrons to form oxygen molecule and water.
(b) Insert glowing wooden splinter into the test tube. Glowing splinter rekindles.
(c) - Blue colour of copper(II) sulphate solution become paler in Set I but blue colour of
copper(II) sulphate solution remain unchange Set II..
- Concentration of Cu2+ ion decreases in Set I but concentration of Cu2+ ion remain unchange
in Set II.
- Cu2+ ion is discharge to form copper atom at cathode in Set I but in Set II rate of Cu2+ ion
discharge to form copper atom at cathode is the same as the rate of copper atom ionise to
form Cu2+ ion at anode
Enrichment Corner (Page 60)

- Oxidised substance: PbS
- Reduced substance: H2O2
- Oxidation number of sulphur before reaction is -2 and after reaction is +6
- Oxidation number of oxygen before reaction is -1 and after reaction is -2
- PbS undergo oxidation reaction because the oxidation number of sulphur increases from -2 to +6
while H2O2 undergo reduction reaction because the oxidation number of oxygen decreases from -1
to -2
ANSWERS CHAPTER 2
Activity 2A (page 66)
1 a) Organic
b) In organic
c) In organic
d) Organic
e) In organic
2 a) Carbon
Compounds containing the element carbon.
compounds
b) Saturated
Hydrocarbons possessing only single bonds. hydrocarbon
c) Organic compounds containing only hydrogen and carbon. Hydrocarbon

d) Organic
Carbon compounds derived from living things.
Compound
e) Hydrocarbons containing at least one double or triple bond between carbon
Unsaturated
atoms
hydrocarbon
Laboratory activity 2A (page 69)
1. Some porcelain chips are used for uniform heating of petroleum and to avoid bumping of the
liquid due to uneven heating.
2. The boiling points of petroleum fractions are in the range of 30̊C - 200̊C. The maximum
temperature that a normal thermometer can record is 110̊C. Thus petroleum fractions which have
a boiling point exceeding 110̊C can not be separated.
3. a) The higher the boiling point, the darker the color of the fractions
b) The higher the boiling point, the higher the viscosity of the fractions
c) The higher the boiling point, the higher the quantity of soot formed after combustion
4. Fraction 1
Self asses 2.1 (page 70)

1 a) Carbon compounds are compounds that contain carbon as their constituent elements
b) - Organic compounds containing only hydrogen and carbon
- Organic compounds containing carbon and hydrogen as well as such as oxygen,
nitrogen, phosphorus or halogen.
c) - Hydrocarbons containing only single bonds between carbon atoms
- Hydrocarbons containing at least one double or triple bond between
carbon atoms
2 a) Cracking is the process of breaking long chain hydrocarbons into smaller hydrocarbons
at a high temperature with the presence of catalyst.
b) (i) C10H22 → C6H14 + C4H8
(ii) C11H26 → C4H8 + C3H6 + C4H12
c)
- The demand for small size hydrocarbons is higher .
- The separation of petroleum to its fractions by fractional distillation does not meet
the demand for small-sized hydrocarbons..
- The cracking process produces smaller sized hydrocarbons that can be used as fuel
as well as raw materials in the petrochemical industry
1
Activity 2F (page 77)
1. The first six members of alcohol

n Molecular formula Number of carbon atom Structural formula Name
1 CH3OH 1 Methanol
2 C2H5 OH 2 Ethanol
3 C3H5 OH 3 Propanol
4 C4H9 OH Buthanol
4
5 C5H11OH 5 Penthanol
6 C6H13 OH 6 Hexanol
The first six members carboxylic acid

n Molecular formula Number of carbon atom Structural formula Name
O
1 || Methanoic acid
0 HCOOH
H – C–OH
H O
| ||
1 CH3COOH 2 H –C –C–OH Ethanoic acid
|
H
H H O
| | ||
2 C2H5COOH 3 H –C–C –C–OH Propanoic acid
| |
H H
H H H O
| | | ||
3 C3H7COOH 4 H –C–C–C –C–OH Butanoic acid
| | |
H H H
H H H H O
| | | | ||
4 C4H9COOH 5 H – C– C–C–C –C–OH Pentanoic acid
| | | |
H H H H
H H H H H O
| | | | | ||
5 C5H11COOH 6 H– C– C– C–C–C –C–OH Hexanoic acid
| | | | |
H H H H H
2
2. Based on the general formula of carboxylic acid CnH 2n+1COOH, there is one carbon atom in
the functional group -COOH. As the first member of a carboxylic acid having one carbon atom,
the value of n in the general formula must start with a value of 0.
3. The carboxyl functional group, always at the end of the carbon chain because the
carbon in the group has formed 3 covalent bonds with O and OH.
Activity 2G (page 78)

1
(i) The members in the homologous series of alkanes and alkenes are represented by the same general
2
formula namely:
a) C2H2n+2 for alkanes
b) C2H2n for alkene
(ii)Consecutive members in the homologous series of alkanes and alkanes are different from one carbon
atom and two hydrogen atoms (CH2) or relative molecular mass = 12
(iii) The physical properties of the members of the homologous series of alkanes and alkenes gradually
change from one member to the next member from gas to liquid.
3
Self asses 2.2 (page 79)
1.
Organic compound
Hydrocarbon Non Hydrocarbon
Alkane Alkene Alkyne Alcohol Carboxylic

acid Ester
CnH2n+2 CnH2n CnH2n-2 CnH2n+1O CnH2n+1 CnH2n+1 COO

n=1,2,3. n=2,3.. n=2,3.. H COOH CmH2m+1 n=1,2,3..
123 012 012
a) (i)
2
P Q R S T
Alkyne
Carboxylic acid Alkene Alcohol Alkane
(ii)
(iii)
P Q R S T
Butanol Butene Propanol Heptane Pentyne
b) i) Substance that exists in the form of gas : Q
ii) Substances that exist in the form of liquid : P,R,S,andT
c) - Boiling point S is higher than Q

- Molecular size S is larger Q
- The van der Waals forces between the molecules S is stronger than Q
- More energy is needed to overcome the attraction between the S molecules compared
to Q molecules
4
Activity 2H (page 82)
a) 2C2H6 + 7O2 → 4CO2 + 6H2O
1
b) C3H8 + 5O2 → 3CO2 + 4H2O
c) 2C6H14 + 19O2 → 12CO2 + 14H2O
d) 2C8H18 + 25O2 → 16CO2 + 18H2O
a) C2H6 + Br2 → C2H5Br + HBr
2
b) Ethane Bromoethane
C2H5Br + Br2 → C2H4Br2 + HBr
Bromoethane Dibromoethane
C2H4Br2 + Br2 → C2H3Br3 + HBr
Dibromoethane Tribromoethane
C2H3Br3 + Br2 → C2H2Br4 + HBr
Tribromoethane Tetrabromoethane
C2H2Br4 + Br2 → C2HBr5 + HBr
Tetrabromoethane Pentabromoethane
C2HBr5 + Br2 → C2Br6 + HBr
Pentabromoethane Hexabromoethane
Experiment 2A (page 87)
1 a) Hexane burns with more soot than hexane
b) % C in hexane = 12(6) x 100% = 83.72%
12(6) + 1(14)
% C in hexene = 12(6) x 100% = 85.71%
12(6) + 1(12)
c) The higher the percentage of carbon by mass per molecule, the more soot is produced by
the flame
- Acidic solution of potassium manganate (VII)
2 a)
- Air bromine
- Hexane does not decolorize the brown color of the bromine water but the hexane
decolorizes the brown bromine water.
- Hexane does not decolorize the purple color of acidic potassium manganate (VII)
solution but hexene decolorize the purple color of acidic potassium manganate (VII)
solution
b) - Hexane is a saturated hydrocarbon that contains a single covalent bond between
carbon atoms. The addition reaction does not occur when an acidic solution of
potassium manganate (VII) is added and
- Hexene is an unsaturated hydrocarbon that contains a double double covalent bond
between carbon atoms, -C = C-. The addition reaction occurs when an acidic solution
of potassium manganate (VII) is produced to produce hexanadiol.
- Balanced equation :
C6H12 +H2O+ 2[O] → C6H12(OH)2
Hexene Hexanadiol
3 When bromine water is dropped into a liquid hydrocarbon, the brown color of bromine
water decolorised // When an acidic solution of potassium manganate (VII) is dropped into
the hydrocarbon, the purple color of potassium manganate (VII) decolorised
5
Activity 2I page 88)
1 - Addition of hydrogen:
C2H4 + H2 → C2H6
- Addition of halogen, Br2
C2H4 + Br2 → C2H4Br2
- Addition of hydrogen halide, HBr
C2H4 + HBr → C2H5Br
- Addition of water (Hydration)
C2H4 + H2 O H3PO4 C2H5OH

300°C/60 atm
- Addition of acidified potasium manganate(VII)
C2H4(g) + H2O [O] C2H4(OH)2

- Addition polymerasation:
H H H H
n C C C C
H H H H
n
where n is any very large integer value
2 a) - Hexene burns more soot than hexane

- % C in hexane = 12(6) x 100% = 83.72%
12(6) + 1(14)
- % C in hexene = 12(6) x 100% = 85.71%
12(6) + 1(12)
- The percentage of carbon by mass in hexene is higher than hexane
1. Put 2 cm3 hexane into the test tube.
b)
2. Add 2 - 3 drops of bromine in 1,1,1-triklotroetana to hexane.
3. Shake the mixture
4. Record all observations
5. Steps 1 to 4 are repeated using hexane to replace of hexene
Observation:
- Hexene decolorises the brown color of bromine water.
- Hexane does not decolorises the brown of color of the bromine water
Laboratory activity 2B (page 89)
1 Yeast contains enzymes that act as catalysts to break down glucose into ethanol and carbon
dioxide .
2 To ensure that the gas released during fermentation is passed through the lime water.
3 Carbon dioxide gas
4 Ethanol
5 - The filtrate is a mixture of ethanol and water that has different boiling points. .The boiling
point of ethanol is lower than that of water.
- When the mixture is heated, ethanol boils first before water at the temperature of 78 ° C
6
- The ethanol vapor formed at its boiling point will be condensed in a Liebig condenser and then
collected in a test tube.
6 Yeast
C6H12O6(ce) 2C2H5 OH (ce) + 2CO2 (g)
Glucose Ethanol
7 Yeast contains the enzyme zymase which can break down carbohydrate or sucrose and fructose
molecules found in fruits into glucose and then to ethanol.
Laboratory activity 2C page 92

A:Oxidation of ethanol
1 Ethanoic acid
2 1. Potassium dichromate(VI) solution
2. Potassium manganate (VII)
3 C2H5OH + 2[O] → CH3COOH + H2O

Ethanol Ethanoic acid Water
4 Acidic
B : Dehydration of ethanol
1 Ethene
2 As a catalyst
3 C2H5OH Porcelain chips C2H4 + H2O.
Ethanol Ethene Water
∆
Activity 2J page 94
1 a) 2CH3OH + 3O2 → 2CO2 + 4H2O

b) 2C3H7OH + 9O2 → 6CO2 + 8H2O
2 a) C4H9OH + 2[O] → C3H7COOH + H2O
Butanol Butanoic acid Water
b) C5H11OH + 2[O] → C4H9COOH + H2O
Pentanol Pentanoic acid Water
3 a) C4H9OH Porcelain chips C4H8 + H2O.
Butanol Butene Water
b) C5H11OH Porcelain chips ∆ C5H10 + H2O.
Pentanol Pentene Water
Laboratory activity 2D page 96
1 Hydrogen Gas
2 a) Magnesium ethanoate
7
b) Sodium ethanoate
c) Copper(II) ethanoate
2CH3COOH + Mg → (CH3COO)2 Mg + H2
a)
3 b) 2CH3COOH + Na2CO3 → 2CH3COONa + H2O + CO2
c) 2CH3COOH + CuO → (CH3COO)2 Cu + H2O
Activity 2K page 99
(i) Molecular formula of alcohol component :CH3OH

Molecular formula of carboxylic acid component: HCOOH
Structural formula of ester :
Methyl methanoate
(ii) Molecular formula of alcohol component:C2H5OH

Molecular formula of carboxylic acid component: C2H5 COOH
Structural formula of ester
Ethyl propanoate
(iii) Molecular formula of alcohol component:C3H7OH

Molecular formula of carboxylic acid component: CH3COOH
Structural formula of ester
Propyl ethanoate
Laboratory activity 2E page 100

1 Esterification
2 Ethyl ethanoate
3 Ethyl ethanoate is less dense than water
4 As a catalyst
5 CH3COOH + C2H5OH → CH3COOC2H5 + H2O
8
Self asses 2.3 page 100
1 a) (i) - - Q burns more soot than P

- - Percentage of carbon by mass in molecules P and Q:
% C in P = 12(3) x 100% = %
12(3) + 1(8)
% C in Q= 12(3) x 100% = %
12(3) + 1(6)
- The percentage of carbon by mass in molecule Q is higher. Therefore Q burns with
more soot than P
(ii) - - P is a saturated hydrocarbon that has only a single bond between carbon atoms.
Addition reaction does not occur between compound P with bromine water.
- - Q is an unsaturated hydrocarbon that has a double bond between carbon atoms.
Addition reaction occurs between molecules Q with bromine water:
- C3H6 + Br2→ C3H6 Br2
b) 2C2H5OH + 7O2 → 4CO2 + 6H2O
Mol of R = 2.3g = 0.05 mol

46gmol -1
From the equation : 2 mol C2H5OH : 4 mol CO2

0.05 mol C2H5OH : 0.10mol CO2
Volume of CO2 = 0.10 mol x 24 dm3mol-1
= 2.4 dm3
c) -R dan T
Ethyl ethanote
d) Acid X is concentrated sulphuric acid
Chemical equation:
2H2SO4 + CaCO3 →CaSO4 + H2O + CO2
Activity 2N page 105

1 C5H11OH
Pentan-1-ol Pentan-2-ol Pentan-3-ol
3-methylbutan-1-ol 2-methylbutan-1-ol 2-methylbutan-2-ol
9
3-methylbutan-2-ol 2,2-dimethylpropan-1-ol
Self assess 2.4 page109
1 a)
2,3-dimethylbutane
b)
3,4-dimethylpent-1-ene
c)
2-ethyl-3-methylhexane
d)
4-methylhex-1-yne
e)
3-methylpentan-2-ol
f)
1,2-dibromohexane
2 i) X: 2-methylbut-2-ene
Y : Pent-2-ene
Z: 2-methylbut-1-ene
ii) - X and Z are isomers
- X and Z have different structural formulae but the same molecular formula which is
C5H10
Achievement Test page 111
1 a) - Petroleum is a mixture of simple or long chain hydrocarbon

- Fractions in petroleum can be separated because each hydrocarbon fraction has its
own boiling point.
- .
10
b) (i) B
(ii) C
(iii) E
c) (i) - The X process is the cracking process
- The catalyst is a mixture of silicon (IV) oxide and aluminum oxide
(ii)
- The cracking process produces smaller sized hydrocarbons that can be used as fuel
as well as raw materials in the petrochemical industry.
2 a) (i)
Hydrocarbons that have at least one double or triple bond between carbon atoms.
(ii)
Alkene
(iii)
Pent-1-ene Pent-2-ena
2-methylbut-2-ena 2-methylbut-1-ena 3-methylbut-1-ena

b) (i)
Hydrogen (Hydrogenation)
(ii)
c) (i)
2C5H10 + 15O2 → 10CO2 + 10H2O
(ii) Number of mol X = 14 g
70 gmol-1
= 0.2 mol
From the equation : 2 mol C5H10 : 10 mol CO2
0.2 mol C5H10 : 1.0 mol CO2
Volume of carbon dioxide = 1.0 mol x 24 dm3mol-1
= 24 dm3
3 a) (i) Fermentation
(ii) Ethanol
(iii)
b) C2H5OH + 3 O2 → 3CO2 + 3 H2O

c) (i) Ethene
(ii)
11
d) The purple color decolorises
e (i) Esterification
(ii) Ethyl ethanoate
Enrichment Corner Page 112
Aim: To prepare two different esters using the same alcohol and different carboxylic acids and identify
their odor.
a) Hypothesis: The same alcohol reacts with different carboxylic acids will produce different ester odors
b) Manipulated variables: ethanoic acid and butanoic acid (carboxylic acid type)
Responding variable: Odor of ester
Constant variable: Pentanol (type of alcohol)
c) Materials and apparatus:
Materials: pentanol, glacial ethanoic acid, butanoic acid, concentrated sulfuric acid, water
Apparatus: Boiling tube, measuring cylinder, test tube holder, dropper, beaker, Bunsen burner
d) Procedure:
1. Put 2 cm3 of glacial ethanoic acid into the boiling tube.
2. Add 2 cm3 of pentanol to the boiling tube containing glacial ethanoic acid.
3. Add 5 drops of concentrated sulfuric acid by using a dropper to the mixture and shake the
boiling tube.
4. Heat the mixture slowly over low heat for two minutes.
5. Pour the contents of the boiling tube into a beaker containing half full water.
6. Record the odor produced
7. Repeat steps 1 to 6 by replacing glacial ethanoic acid with butanoic acid.
a) Observation:
Carboxylic acid Alcohol Odor

Ethanoic acid Pentanol
Butanoic Pentanol
b) Chemical equation:
- Reaction between ethanoic acid and pentanol
CH3COOH + C5H11OH → CH3COO C5H11 + H2O
Ethanoic Pentanol Pentyl ethanoate Air
- Reaction between butanoic acid and pentanol

C3H7COOH + C5H11OH → C3H7COOC5H11 + H2O
Butanoic Pentanol Pentyl butanoate Air
12
ANSWERS CHAPTER 3
Laboratory Activity 3A (page 117)

Results:
Reactants Initial Temperatue (0C) Highest or Lowest
Temperature (0C)
Solid sodium hydroxide, NaOH +
water
Solid ammonium nitrate, NH4NO3

+ water
Solid sodium thiosulphate,

Na2S2O3 + water
Solid anhydrous calcium chloride,

CaCl2 + water
Discussions
1. (a) Solid ammonium nitrate, solid sodium thiosulphate, solid anhydrous calcium chloride
(b) Sodium sodium hydroxide.
2.
Exothermic reaction Endothermic reaction
Sodium hydroxide + water Ammonium nitrate + water
Anhydrous calcium chloride + water
Sodium thiosulphate + water
Self Assess 3.1 (page 119)

1. (a)
Energy
CaO + CO2
∆ H = + 540 kJ mol-1
CaCO3
(b)
1 The decomposition of calcium carbonate, CaCO3 is an endothermic reaction.
2 When 1 mole of calcium carbonate, CaCO3 is decomposed , 540 kJ energy is absorbed from the
environment.
3 The energy content of the products is higher than the energy content of the reactant..
2. (a) Exothermic reaction

(b) Reactants (Pb(NO3)2 and Na2SO4) undergo bond breaking while products (PbSO4 and NaNO3)
undergo bond formation.
(c) Energy is absorbed during bond breaking while heat is released during bond formation..
1
(d)
Energy
Pb(NO3)2 + Na2SO4
∆ H = - 50 kJ mol-1
PbSO4 + 2NaNO3
Laboratory Activity 3B (pages 123 and 124)

Results:
Set 1 Set II
Temperature (0C) Silver nitrate, AgNO3 Magnesium nitrate, MgNO3)2
+ sodium chloride, NaCl + sodium carbonate, Na2CO3
Initial temperature of nitrate salts
Initial temperature of sodium salts
Average temperature of both
solutions
Highest temperature
Increase in temperature, θ
Discussions :
1. Double decomposition reaction (Precipitation reaction)
2. Steps in calculations:
(a) Heat of precipitation of silver chloride
(i) Calculate the number of moles or silver chloride, AgCl precipitate formed
Number of moles of silver ions, Ag+ = number of moles of silver nitrate, AgNO3 solution
25
= 0.5 mol dm-3 x dm3 = 0.0125 mol
1000
Number of moles of chloride ions , Cl = number of moles of sodium chloride, NaCl solution
-
25
= 0.5 mol dm-3 x dm3 = 0.0125 mol
1000
Ionic equation : Ag (aq)
+
+ Cl (aq)  AgCl(s)
-
From the ionic equation. 1 mole of silver ions, Ag+ reacts with 1 mole of chloride ions, ,Cl-
to produce 1 mole of silver chloride, AgCl.
Therefore, 0.0125 moles of silver ions, Ag+ reacts with 0.0125 moles of chloride ions, Cl- to
produce 0.0125 moles of silver chloride, AgCl.
(ii) Calculate the heat change:

Maas of reacting mixture = Total volume of reacting mixture x density of solution
= (25 + 25) cm3 x 1 g cm−3 = 200 g
Heat released in the reaction Q = mcθ
= 200 g x 4.2 J g-1 0C-1 x θ
=xJ
2
(iii) Calculate the heat change for the formation of 1 mole of precipitate
Precipitation of 0.0125 moles of silver chloride releases x J heat
Therefore the precipitation of 1 mole of silver chloride releases = xJx
1 𝑚𝑚𝑚𝑚𝑙𝑙
0.0125 𝑚𝑚𝑚𝑚𝑚𝑚
= 80x J heat
= 0.08x kJ heat
(iv) Write the heat of precipitation by putting a negative sign for an exorhermic reaction
Heat of precipitation of silver chloride = - 0.08 x kJ mol-1
Note: Follow the same steps to calculate the heat of precipitation of magnesium carbonate.
3. The thermochemical equation for the heat of precipitation of silver chloride:

AgNO3 + NaCl  AgCl + NaNO3 ∆H = - 0.08x kJ mol-1
4. The thermochemical equation for the heat of precipitation of magnesium carbonate:

Mg(NO3)2 + Na2CO3  MgCO3 + 2NaNO3 ∆H = - (Nilai kiraan) kJ mol-1
5. The energy level diagram for the heat of displacement of silver chloride.
Heat
AgNO3(aa) + NaCl(aak)
∆ H = - 0.08x kJ
AgCl(s) + NaNO3(aq)
The energy level diagram for the heat of displacement of .magnesium carbonate.
Energy
Mg(NO3)2+ Na2CO3
∆ H = - …. kJ mol-1 (Write the calculated value)
MgCO3 + 2NaNO3
6. Different.
Difference due to:
(i) Heat is lost to the surroundingd.
(ii) Polystyrene cup absorbs heat.
Laboratory Activity 3C (pages 126 and 127)

Procedure::
1. Measure 25 cm3 of 0.2 mol dm−3 copper(II) sulphate, CuSO4 solution and pour it into a
polystyrene cup.
2. Put a thermometer into the solution and leave aside for two minutes.
3. Record the initial temperature of the solution in a table..
3
4. Add one spatula od magnesium powder, Mg quickly and carefully into the polystyrene cup..
5. Close the polystyrene cup and stir the mixture using the thermometer.
6. Record the highest temperature of the mixture.
7. Repeat steps 1 to 6 by using zinc powder, Zn to replace magnesium, Mg.
Results:
Metal Magnesium Zinc
Initial temperature of copper(II) sulphate
solution (0C)
Highest tempearute of mixture (0C)
Temperature change, θ (0C)
Discussions:
1. (a) Chemical equation : Mg(s) + CuSO4(aq)  Cu(s) + MgSO4(aq)
Zn(s) + CuSO4(aq)  Cu(p) + ZnSO4(aq)
Ionic equation : Mg(s) + Cu2+ (aq)  Cu(s) + Mg2+(aq)

Zn(s) + Cu2+(aq)  Cu(s) + Mg2+(aq)
(b) Steps in calculations;:

(i) Calculate the number of moles of copper, Cu displaced from copper(II) sulphate, CuSO4
solution
Number of moles of copper(II) sulphate, CuSO4 solution
25
= 0.5 mol dm-3 x dm3 = 0.0125 mol
1000
From the equation, 1 mole of copper, Cu is displaced from 1 copper(II) sulphate, CuSO4
solution.
Therefore 0.0125 moles of copper, Cu is displaced from 0,0125 moles of copper(II)
sulphate, CuSO4 solution.
(ii) Calculate the heat change.

Mass of solution, m = 25 cm3 x 1 g cm3 = 25 g
Heat released in the reaction, Q = mcθ
= 25 g x 4.2 J g-1 0C-1 x θ
= xJ
(iii) calculate the heat of displacement
Displacement of 0.0125 mole of copper, Cu releases x J heat
Therefore, 1 mole of copper, Cu will release = xJx
1 𝑚𝑚𝑚𝑚𝑚𝑚
= y J = z kJ heat
(iv) Write the heat of displacement by putting a negative sign for an exothermic reaction.
Heat of displacement of copper, Cu = - z kJ mol-1
Note: Follow the same steps to calculate the heat of displacement of copper, Cu using zinc,
(c) Magnesium is a more electropositive metal compared to zinc. Therefore the change in
temperature is higher. Thus the value of heat of precipitation is different.
4
Tenaga Tenaga
(d) Mg + CuSO4 Zn + CuSO4
∆ H = - …. kJ mol-1 ∆ H = - …. kJ mol-1
Cu + MgSO4 Cu ) + ZnSO4
Note: Put the caluculated value of ∆H
2. To ensure that all copper(II) ions are completely displace to form copper atoms.
3. (i) Brown solid deposited.

( ii) The blue colour of the solution becomes pale blue
4. Operational definition of heat displacement : When magnesium metal is added to copper (II)
sulphate solution to displace 1 mole of copper, the thermometer reading increases.
5. (i) Stirring the solution slowly and continuously through out the experiment to ensure the
temperature of the mixture is uniform.
(ii) The solid metal is added quickly and carefully,
(iii) Using metal powder and not granules to ensure the reaction occurs faster,
6. Same. Heat of displacement only involves copper ions which are present in in copperII) nitrate
solution. .
Activity 3A (page 127)

(a) Cu + 2AgNO3  2Ag + Cu(NO3)2 ∆H = - 105 kJ mol-1
100
(b) Number of moles of silver ions = 0.5 mol dm-3 x dm3 = 0.05 mol
1000
From the thermochemical equation, 1 mole silver displaced releases 105 kJ heat
Therefore 0.05 moles of silver displaced releases 0.05 mol x 105 kJ mol-1 = 5.25 kJ heat
Q
(c) Q = mcθ there fore, θ =
mc
5.25 x 1000
θ= = 12.5 0C
100 x 4.2
(d)
Heat
Cu + 2 AgNO3
∆ H = - 105 kJ mol-
2Ag+ Cu(NO3)2
5
100
(e) ) Number of moles of silver nitrate solution = 1.0 mol dm-3 x dm3 = 0.1 mol
1000
From the thermochemistry equation, displacement of 1 mol of silver released 105 kJ of heat
There fore 0.1 mole of silver displaced releases 0.1 mol x 105 kJ mol-1 = 10.5 kJ of heat
Q
Q= mcθ there fore, θ =
mc
10.5 x 1000
θ= = 25 0C
100 x 4.2
The number of moles of solution is twice, thus the change in temperature is also twice.
Experiment 3A (pages 128 and 129)

Hypothesis::
The reaction between a strong acid and weak alkali will produce a higher heat of neutralisation as
compared to the reaction between a weak acid and weak alkali.
Discussions:
1. A: HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
CH3COOH(aq) + NaOH(aq)  CH3COONa(aq) + H2O(l)
B: HCl(aq) + NH3(aq) )  NH4Cl(aq)

CH3COOH(aq) + NH3(aq  CH3COONH4(aq)
Note: Explanation of the neutralisation reaction using aqueous ammonia.

1. Chemical equation : HCl(aq) + NH3(aq)  NH4Cl(aq)
(NH4OH molecules do not exist)
2. Neutralisation of acids with alkalis will produce salt and water but in the case of aqueous
ammonia, water in the equation is not shown.
3. Aqueous ammonia, NH3 is a weak alkali that ionises partially in water to produce hydroxide
ions, OH-.
NH3(aq) + H2O(l) ⇋ NH4+(aq) + OH-(aq)
4. The neutralisation reaction between an acid with aqueous ammonia also produces water and
this can be shown in the ionic equation.
5. H+ ions from the acid reacts with OH- ions to form ammonia to produce water.
Ionic equation: H+(aq) + OH- (aq)  H2O(l)
2. From the equation, 1 mole of acid reacts with 1 mole of alkali.

Steps in calculating each reaction:
(i) Calculate the number of moles of water produced.
Number of moles of H+ ions = Number of moles of acid
50
= 1.0 mol dm-3 x dm3 = 0.05 mol of H+ ions
1000
Number of moles of OH- ions = number of moles of alkali
50
= 1.0 mol dm-3 x dm3 = 0.05 mol of OH-ions
1000
6
There fore , 0.05 moles of H+ ions reacts with 0.05 moles of OH- ions to produce 0.05 moles
of water.
(ii) Calculate the heat change:

Total volume of solutions = volume of acid + volume of alkali
= (50 + 50) cm3 = 100 cm3
Mass of mixture, m = 100 cm3 x 1 g cm3 = 100 g
Change in temperature of the mixture, θ
Heat released in the reaction, Q = mcθ
= 100 g x 4.2 J g-1 0C-1 x θ = p J
(ii) Calculate the heat of neutralisation:

0.05 moles of water formed produces p J heat
There fore, 1 mole water formed produces
pJx = q J = r kJ heat
(iv) Write the heat of neutralisation for the reaction between an acid and an alkali is – r kJ mol-1
Note : Negative sign shows that the neutralisation reaction is exothermic
3. Construct the energy level diagram as follows:

Note: Write the chemical formulae for reactants and products
Heat
Reactants
∆ H = - …. kJ mol-1
Products
4. The heat value of neutralisation between a strong acid and a strong alkali is higher than the heat of
neutralisation when using a strong acid with a weak alkali.
The value of the heat of neutralisation using a strong alkali with a weak acid is also lower than
the heat of neutralisation using a strong acid with a strong alkali..
The value of heat of neutralisation is the lowest for the reaction between a weak acid and a weak
alkali
5.
• Hydrochloric acid is a strong acid and sodium hydroxide solution is a strong alkali. Both
solutions ionise completely in water to produce a high concentration of ions.
• The neutralisation reaction involves only the combination of hydrogen ions and hydroxide
ions to form water molecules.
• The heat released is not reabsorbed to ionise strong acids or strong alkalis
• Ethanoic acid is a weak acid and ionises partially in water. The concentration of hydrogen
ions is low because most weak acids exist as molecules.
• During neutralisation, a small amount of heat released during the formation of 1 mole of
water molecules is reabsorbed by the ethanoic acid molecule to ionise the acid molecules
completely.
• Thus, when the released decreases, the heat of neutralisation also decreases.. The same goes
for aqueous ammonia solution which is a weak alkali.
7
6. The value of heat of neutralisation obtained in this experiment is lower. This is due to:
a. part of the heat is released to the surroundings.
b. the polystyrene cup absorbs heat.
Activity 3B (page 132)

(a) Mass. m = (100 + 100) cm3 x 1 g cm-3 = 200 g
Temperature change, θ = (43.5 – 30.0) 0C = 13.5 0C
Q = 2000 g x 4.2 J g-1 0C-1 x 13.5 0C = 11340 J
Number of moles of water = Number moles of hydrogen ions or number of moles of hydroxide
ions
100
= 2.0 mol dm-3 x dm3 = 0.2 mol
1000
Neutralisation of 0.2 moles of hydrogen ions with 0.2 moles of hydroxides ions releases
11340 J heat
There fore the heat released when I mole of hydrogen ions reacts with 1 mole of hydroxide ions
releases
= 11340 J x = 5670 J = 56.7 kJ
Heat of neutralisation = – 56.7 kJ mol-1
(b) HCl + NaOH  NaCl + H2O ∆H = - 56.7 kJ mol-1
(c) Heat
HCl + NaOH
∆H = - 56.7 kJ mol-1)
NaCl + H2O
(d) Temperature change is 13.5 0C. Nitric acid is strong acid and ionises completely in water.
Experiment 3C (page 133)

Results:
Alcohol Methanol Ehtanol Propanol Butanol
Initial temperature of water (0C) T1
Highest temperature of water (0C) T2
Increase in temperature (0C) (T2 -T1) = θ
Mass of lamp before burning (g) m1
Mass of lamp after burning (g) m2
Mass of alcohol burnt (g) m1 -m2 = ma
Note: Unknown values are given to show steps of calculations.
Discussions:
1. (a) Example : Heat of combustion of methanol, CH3OH
Molar mass of methanol, CH3OH =12 + 4(1) + 16 = 32 g mol-1
𝑚𝑚a
Number of moles of methanol, CH3OH burnt = mol = p mol
32
8
Heat released during thhe combustion of methanol, CH3OH = heat absorbed by water
= mcθ
= 200 g x 4.2 J g-1 0C-1 x θ = q J
Combustion of p mol of methanol, CH3OH releases q J haba
There fore, the combustion of 1 mole of methanol. CH3OH releases = q J x = r J heat
𝑥𝑥 𝑚𝑚𝑚𝑚𝑚𝑚
= s kJ heat
The heat of combustion of methanol, CH3OH = – s kJ mol-1
Using the same method, calculate the heat of combustion of ethanol, propanol and butanol.
(b) The thermochemical equation for the heat of combustion of methanol:

3
CH3OH + O2  CO2 + 2H2O ∆H = - s kJ mol-1
2
The thermochemical equation for the heat of combustion of ethanol :

C2H5OH + 3O2  2CO2 + 3H2O ∆H = - ...... kJ mol-1
The thermochemical equation for the heat of combustion of propanol:

9
C3H7OH + O2  3CO2 + 4H2O ∆H = - .... kJ mol-1
2
The thermochemical equation for the heat of combustion of butanol:

C4H9OH + 6O2  4CO2 + 5H2O ∆H = - .... kJ mol-1
Note: Write the value of ∆H calculated for each alcohol.
(c) Energy level diagram for the combustion of alcohol

Example: methanol
Energy
3
CH3OH + O2
2
∆H = - s kJ mol-1
3CO2 + 4H2O
Note: Construct the energy level diagram for other alcohols
2. As the number of carbon atoms per molecule in the alcohol increases, the value of heat of
combustion also increases.
3. Operational definition of heat of combustion of alcohol: Temperature rise when 1 mole of
alcohol is burned completely.
4. Copper is a good conductor of heat.
5. The bottom of the container turns black. The substance is soot (carbon) due to the incomplete
combustion of alcohol.
9
6. Precautions:
(i) Use a windshield
(ii) Make sure the flame of the lamp touches the bottom of the copper container.
(iii) Stir the water in the copper container continuously.
(iv) When the flame of the alcohol is extinguished, the lamp must be weighed immediately as the
alcohol is very volatile

1. Heat released,, Q = 100 g × 4.2 J g-1 0C-11 × 4 ºC = 1680 J
Ionic equation : Ca2+ + CO32-  CaCO3
From the equation: 1 mole of Ca2+ ions reacts with 1 mole of CO32- ions to produce 1 mole of
CaCO3 precipitate.
Number of moles of Ca2+ ions = Number of moles of CO32- ions.
50
= 2.0 mol dm-3 x dm3 = 0.1 mol
1000
The formation 0.1 moles of CaCO3 releases 16800 J heat.
1
There fore the formation of 1 mole CaCO3 releases 16800 J x = 168000 J = 168 kJ
0.1
Heat of precipitation of CaCO3 = + 168 kJ mol-1

50
2. Number of moles of Fe2+ ions = 0.25 mol dm-3 x dm3 = 0.0125 mol
1000
Mass of solution, m = 50 cm3 x 1 g cm-3 = 50 g
Given, ∆H = 80.6 kJ mol-1
1 mole of Fe2+ ions displaced releases 80.6 kJ of heat.
There fore 0.0125 moles of Fe2+ ions releases 80.6 kJ x 0.0125 = 1.075 kJ heat = 1075 J heat
Q
Q = mcθ, There fore θ =
mc
1075
θ= = 5.12 0C
50 x 4.2
3. Heat of neutralisation using ethanoic acid is less then using hydrochloric acid.
Ethanoic acid is a weak acid while hydrochloric acid is a strong acid.
Some of the heat given out is used to completely ionise ethanoic acid.
Calculation of Heat of Combution Activity in QR Code (page 135)

0.37
1. (a) Number of moles of butanol, C4H9OH = mol = 0.005 mol
74
Heat released,, Q = 0.005 mol x 2671 kJ mol = 13.355 kJ
-1
Q
(b) Q = mcθ, there fore θ =
mC
Mass of water = 100 cm3 x 1 g cm-3 = 100 g
θ= 13355 J
100 g x 42 J g-1 oC-1
= 31.8 0C
2. Calculate the number of moles of methanol, CH3OH.

Molar mass of methanol, CH3OH = 12 + 4(1) + 16 = 32 g mol-1
Mass of methanol, CH3OH used = (145.91 – 144.95) g = 0.96 g
10
Number of moles of methanol, CH3OH = mass of methanol
molar mass of methanol
= 0. 96 g = 0.03 mol
32 g mol -1
Calculate the heat released.

Mass of water, m = 200 cm x 1 g cm-3 = 200 g
Temperature change, θ = (59.0 – 29.0) 0C = 30.0 0C
Heat released, Q = mcθ

= 200 g x 42 J g-1 oC-1 x 30.0 0C = 25200 J =25.2 kJ
Calculate the heat of combustion of methanol, CH3OH
Combustion of 0.03 moles of ethanol, CH3OH releases 25.2 kJ heat
25.2 kJ
There fore, 1.0 mol of ethanol releases = 840 kJ mol-1
0.03 mol
Heat of combustion of methanol, CH3OH = - 840 kJ mol-1
3. Calculate the heat released

Mass of water, m = 2150 cm3 x 1 g cm-3 = 2150 g
Temperature change, θ = (100.0 - 25.0) 0C = 75.0 0C
Heat released, Q = 2150 cm3 x x 42 J g-1 oC-1 x 75.0 0C = 677 25 kJ
= 677.25 kJ
Calculate the mass of hexane, C6H14
Molar mass of hexane, C6H14 = 6(12) + 14(1) = 86 g mol-1
35000 kJ of heat is relapsed by 86 g hexane, C6H14
86 g x 667.25 kJ
There fore the mass of hexane, C6H14 needed to release 677.25 kJ = = 1.53 g
35000 kJ

2. The most suitable fuel for frying eggs or making pop corn is natural gas.
3. Justification of selection:
• Fuel value of natural gas is higher than other fuels.
• Burning 1 gram of natural gas will produce 50 kJ of heat energy.
• Although the fuel value of hydrogen is the highest, the handling of hydrogen as fuel for cooking
is not suitable because the storage of hydrogen gas as hydrogen liquid is difficult.
• Furthermore hydrogen gas is flammable and forms an explosive mixture with air.
• Natural gas contains propane and butane which can be compressed under pressure and filled
into cylinders of various sizes. This facilitates the storage and transportation of natural gas.
• The burning of natural gas in sufficient oxygen for cooking does not produce soot or smoke. So,
it is clean and environmentally friendly.

1. The fuel value of a fuel is the amount of heat energy given out when one gram of the fuel is
completely burnt in excess oxygen..
2. (a) Hydrogen gas

(b) The combustion of hydrogen is clean as it will only produce water. Incomplete combustion of
gasoline can produce soot, toxic carbon monoxide gas and carbon dioxide gas. This will
increase air pollution.
11
Achievement Test (page 146)
1. m = 200 g θ = 7 0C
Q = 200 g x 42 J g-1 oC-1 x 7 0C = 5880 J
2. (a) (i) Exothermic reaction. ∆H is negative

(ii) 2H2 + O2  2H2O ∆H = -282 kJ mol-1
(iii)
Energy
2H2 + O2
∆H = -282 kJ mol-1
2H2O
50
(b) (i) Number of moles of hydrogen = = 25 mol
2
(ii) 1 mole of hydrogen releases 282 kJ heat.
There fore, 25 moles of hydrogen releases 25 mol x 282 kJ mol-1 = 7050 kJ
(c) Hydrogen can be used as fuel because the energy released when 1 g of hydrogen is burned is very
large. It is also a clean fuel because its combustion only produces water. Hydrogen fuels are also
renewable energy.
Hydrogen, on the other hand, is a flammable gas and its storage is difficult. The cost of
producing hydrogen fuel is also expensive.
Enrichment Corner (page 140)

Sodium metal is a highly reactive metal and its reaction is very exothermic. Sodium metal is not as
easy to handle as other metals because it does not exist freely and it only exist as compounds.
Therefore, sodium metal is not suitable for use as a hot pack.
Additional Exercises in QR Code (page 140

1. Number moles of Pb2+ ions = Number of moles of SO42- ions
= 50 x 2.0 = 0.1 mol
1000
Let the heat released is Q J
There fore, heat Q = (50 +50) cm3 x 42 J g-1 oC-1 x 10 0C ................Equation 1
(a) Number moles of Pb2+ ions = Number of moles of SO42- ions

= 100 x 2.0 = 0.2 mol
1000
The number of moles is double, there fore the heat released will also be double.
2Q = (100 + 100) cm3 x 42 J g-1 oC-1 x θ ...............Equation 2
12
Comparing equation 1 and 2
There fore, θ = 10 0C [Does not change]
(b) Concentration of both solutions is halved

Number of Pb2+ ions = Number of SO42- ions
= 50 x 1.0 = 0.05
1000
The amount of substance is half, ther fore there fore the heat released is also half
1
Q = (50 + 50)cm3 x 42 J g-1 oC-1 x θ .............Equation 3
2
Compare equation 1 and 3
There fore , θ =5 0C [Half]
(c) Concentration of the solution are not the same

Number of moles of Pb2+ ions = 50 x 2.0 = 0.1 mo
1000l
Number of moles of SO4 ions = 50 x 1.0 = 0.05 mol
2-
1000
The number of Pb ions is in excess, there fore the calculation will be based on the number of
2+
moles of SO42- ions.

The number of moles of SO42-_ is half, there fore the heat released is also half, and the rise in
temperature will be half which is 5 0C.
(d) Conclusion : The temperature rise depends on the number of moles of ions used.
• The number of moles of ions depends on the volume and concentration of reactants.
• When the moles of ions are the same, the temperature does not change.
• When the number of moles of ions is halved, the temperature rise is also halved.
• Reactant with excess number of moles of ions is not taken into account in the calculation.
(also known as limiting factor)
2. Q = 200 cm3 x 42 J g-1 oC-1 x 50 0C = 42000 J = 42 kJ

1376 kJ of heat is released by 46 g of ethanol
There fore 42 kJ of heat is released by 42 kJ x 46 g mol-1 = 1.40 g
1376 kJ mol-1
13
ANSWER CHAPTER 4
Test Yourself 4.1 (Page 150)
1. (a) One type

(b)
Prop-1-ene
2. – Polymer can be used as chemical storage container as it is resistance to chemicals and heat.
- Polymer such as PVC can be used as water pipe because more durable and not easily broken.
- As synthetic polymer last longer, it needs to be disposed of properly so that it will not cause
environmental pollution.
(i) 2-metylbut-1,3-diene
(ii) Protein membrane on rubber particle is negatively charged which causes rubber particles
to repel one another.
(iii) Characteristics of natural rubber
a. Soft solid at room temperature for easy usage.
b. Elastic to allow gloves to be easily worn.
c. Water resistant so that hands are protected from external elements.
Activityi 4E (Page 155)
(a) Coagulant: Formic acid, formaldehyde, vinegar.

Anticoagulant: Ammonia solution, sodium hydroxide solution.
(b) Natural coagulation process is due secretion of lactic acid by bacteria. Lactic acid is secreted
in a very small quantity and cause the coagulation to take place very slowly.
(c)
-Negatively charged protein membrane - Bacteria in the air enters the latex and
causes rubber particle to repel one secret lactic acid.
another as they are brought close. -Lactic acid contains hydrogen ion, H+
which is positively charged and able to
neutralise negatively charged protein
membrane.
- Anticoagulant such as ammonia - Rubber protein membranes remain
solution is added into the latex negatively charged.
- Anticoagulant contains hydroxide - Latex is prevented from coagulating.
ion, OH- which is negatively
charged.
- Hydroxide ion, OH- will
neutralise the positively charged
hydrogen ion, H+.
Laboratory Activity 4C (Page 156)
Discussion
1. Ethanoic acid is a coagulant to coagulate the latex.
2. To obtain even thickness.
3. Rubber strips produces are harder and can withstand high temperature.
Selff Asesses 4.2 (Page 158)

1. Polyisoprene
2. (a) The presence of hydrogen ions, H+ in the acid neutralise the negatively charged protein
membrane.
(b) - Hydrogen ions, H+ in the acid neutralise the negatively charged protein membrane.
- Rubber particles collide with one another that leads the protein membrane to break.
- Rubber polymers entangle causes latex to coagulate.
3. Free from sulphur and chemical allergies.
4. Similarity: Vulcanised and unvulcanised rubber are both originated from latex.
Characteristic Vulcanised rubber Unvulcanised rubber

Elasticity More elastic Less elactic
Hardness Hard Soft
Strength High Low
Resistance to heat Resistance to high heat Less resistance to high heat
Resistance to oxidation More resistant to oxidation More prone to oxidation

1. A polymer that is elastomeric in nature or elastic polymer.

2. (a) Nitrile rubber
(b) More resistance to oil and solvent.
- Synthetic runner is more resistant towards heat and abrasion to allow products such as tyres
to be manufactured.
- Synthetic rubber is hard to decompose naturally and can cause pollution.
- Synthetic rubber needs to be disposed of systematically so that pollution can be prevented.
Achievement Test (Page 163)
1. (a)
(b) Addition polymerisation.

(c) Production of plastic bottles.
(d) Polymer has higher boiling point than its monomer due to high intermolecular force of
attraction between polymer chain.
2. (a) Natural polymer: polyisoprene.

Synthetic polymer: Polyvinyl chloride.
(b) Natural polymer can be found in nature while synthetic polymer is manmade through
chemical reaction.
(c) (i) Thermosetting polymer is a polymer that cannot be remoulded after heating.
(ii) Hard to decompose naturally difficult to be disposed of.
3. (a) Rubber strip M because natural rubber is less elastic.

(b) Rubber strip M.
Enrichment corner
- Polymerisation process involved is condensation polymerisation.

- Water molecule is produced.
H2O
ANSWERS BAB 5

1. Unsaturated fats are fats that contain unsaturated fatty acids.
2. Ester.
3. Similarity: Consists of the same homologous series.
Differences:
Aspect Fats Oils
Source Animals Vegetables
Physical state at Solid Liquid
room temperature
Melting point High Low
Type of fatty acid High percentage of High percentage of
saturated fatty unsaturated fatty
acids acids
4. Use of oils and fats:

a. As biofuel.
b. Source of nutrition
c. Production of soap and personal care.
d. Manufacture of animal feeds.
Experiment 5A (page 175)

Hypothesis : Detergent is more effective than soap in hard water.
Variables :
Manipulated variable : Soap and detergent
Responding variable : Effectiveness of cleaning
Fixed variable: Volume of hard water
Procedure:
1. Pour 50 cm3 hard water into a beaker.
2. Weigh 5 g of soap powder and pour it into the beaker.
3. Put a piece of cloth with greasy stain into the beaker.
4. Clean/ Scrub / Wash the cloth.
5. Record the observation.
6. Repeat steps 1 to 5 using liquid detergent.
Results:
Type of cleansing agent Observation

Soap
Detergent

1. Soap is a sodium or potassium salt of fatty acids.
2. Detergent is a sodium or potassium salt of sulfonic acid.
3. Saponification.
4. CH3(CH2)16COOH + NaOH  . CH3(CH2)16COONa + H2O
5. Sulphonation and neutralisation.
1
6. Soap bubbles will lower the surface tension of water surface in the following ways:
• Soap bubbles are a layer of spherical soap film that covers air or gas.
• The film consists of a thin layer of water trapped between two layers of soap molecules.
• The hydrophilic part of the soap molecule will dissolve in water while the hydrophobic
hydrocarbon chain part tends to avoid water.
• The hydrophobic ends of soap molecules accumulate on the surface, avoiding water and stays
away from the water molecular layer.
• As a result, water molecules separate from each other. Increased distances between water
molecules will cause a decrease in surface tension, which allows bubbles to form. Thus, soap
can wet the surface of the fabric.

1. Preservatives, emulsifiers, antioxidants, thickeners, flavours, dyes and stabilisers.
2. (a) Antioxidants
(b) (i) Slows down the oxidation of fats in food.
(ii) Prevent oily or greasy food from becoming rancid.
(c) Emulsifiers and dyes / salts

1. Ginger - removes wind in the body.
Aloe vera - treats skin diseases
Tamarind – Juice of the fruit relieves cough
2. Analgesics, antibiotics, psychotic drugs, anti allergies and corticosteroids.
3. Analgesics
4. Make-up cosmetics, treatment cosmetics and fragrances.
.
1. Nanoscience is a study of processing of substances at nano scale while nanotechnology is the
development of substances or gadgets using the properties of nano particles.
2. Nano particle sizes, ranging from 1 to 100 nanometres, enable various applications to be invented
due to its extremely small size. For example health care can be improved with the production of
more effective drugs or devices. In the cosmetic fields, for example, the usage of nano particles is
able to provide more satisfying outcomes as these extremely small particles have the ability to
easily penetrate the skin even more.
3. Examples of applications of nanotechnolgy:

a. Production of semiconductors and electronics:
• Smaller and more efficient semiconductors.
• High conductivity wiring system.
b. Medical:
• Highly sensitive testing device.
• More effective drug delivery system.
c. Energy and electricity:
• Smaller and more efficient solar cell.
• Long-lasting battery.
d. Agriculture:
• More effective pesticide.
• Highly efficient and thorough fertilisation.
2
Textile:
• Water, fire and dirt resistant fabrics.
• Anti-wrinkle and UV protective fabrics.
Food:
• Nano-scale food additives.
• Anti-microbial food packaging.
4. Graphene has a high surface area and cavity size that only allows water molecules to pass
through it.

1. Green Technology is a technology developed to reduce the impact of human activities on the
environment.
2. To prevent groundwater sources from being polluted.
3. Use of Green Technology in wastewater treatment:
a. Use of bacteria to decompose harmful substances.
b. Isolation of solid waste from wastewater treatment
c. Use of bacteria in wastewater treatment.
4. The use of solar energy in vehicles can reduce the release of carbon dioxide gas which
contributes to the greenhouse effect.
Achievement Test (page 196)
1. (a) Unsaturated fats
(b) The double covalent bonds in fatty acids P easily reacts with oxygen in the air and is further
oxidized.
2. (a) (i) Saponification

(ii) Ester
(b) Potassium hydroxide
3. (a) Cleansing agent A : Detergent Cleansing agent B: Soap
(b) Cleansing action of cleansing agent A is more effective in hard water while cleaning agent B
is less effective hard water.
Hard water contains Ca2+ ions and Mg2+ ions.
Anions in cleansing agent B combines with these ions and forms scum which are insoluble
salts.
Cleansing agent A do not form scum.
4. (a) P – Salt Q - Benzoic acid R - Pektin
(b) P – preservatives or flavouring
Function:
Preservative – Prevent or delay the growth of bacteria or fungi to make the food last longer
Flavouring – Replace the loss of original flavours during food processing.
Add flavour or fragrant smell to make food tastier.
Q – preservative
Function - Prevent or delay the growth of bacteria or fungi to make the food last longer.
R – stabiliser
Function – gives a uniform and smooth texture.
5. (a) Medicine X – antiseptic Medicine Y – antibiotic

(b) Drug Y needs to be consumed so that all bacteria are killed. If not consumed, higher doses of
the drug are needed if infected with the same disease in the future.
3
6. Nanotechnology is the development of materials or devices by utilising the characteristics of nano
particles.
(b) Apply cream evenly covering the skin for better protection.
(c) Uses of nanotechnology:
- Production of unmanned vehicles
- Production of electrical goods components.
7. (a) Garbage is disposed using a solvent treatment system that can control the leachate from
contaminating groundwater sources.
(b) Hazardous chemicals can be decomposed first and wastewater treated before discharged into
the river. This allows the a treated and cleaner effluent to flow into the river.
(c) Sludge material can be processed into fertiliser or material for biogas production.
Enrichment Corner (page 197).

- Virgin coconut oil lasts longer
- Low content of unsaturated fatty acids makes virgin coconut oil more difficult to oxidise and last
longer.
Answers to Extra Exercises in QR Code

1. (a) (i) Preservative
(ii) Slows down or prevent the growth of bacteria or fungi so that food can be kept for a longer
period of time.
(b) (i) Natural dye from pandanus leaves.
(ii) Sugar because diabetic patients should not consume too much sugar
(iii) Artificial sweetener for example : stevia / sorbitol / aspartame
2. (i) Analgesic
(ii) Aspirin contains acid which is the -COOH group (carboxylic acid) and is not suitable for
children or gastric patients. as it will cause stomach ulcers.
Codeine does not contain acids.
3. Suitable because traditional medicine does not contain synthetic chemicals, only natural
ingredients.
It is not suitable because if the production does not follow the correct dosage or measurement, it
may not be safe to consume..
[select any one]
4. Steps need to be taken:

- Check if the product is approved by the Ministry of Health.
- If there is no approval, check the content of the product does not contain prohibited substances
such as hydroquinone, tretinoin or mercury.
- If there is a "tester", test the product on the back of the wrist. This is to test whether one is
allergic to the basic ingredients used in the manufacture of the product
5. (a) Medicine:
The field of nanomedicine where the use of nanoparticles has been successfully applied.
Examples are in assisting in the dissolving of drug-active molecules, cargo delivery of drug-
4
active molecules to specific areas of the body that needs treatment, as antibacterial and also in
disease diagnosis.
(b) Energy
Production of a more efficient and smaller solar cell.
Batteries that last longer
(c) Agriculture
Manufacture of fertilisers containing 'nano-activators' or 'nano-stimulants' that act as
stimulants for fruit production.
More effective pesticide production
5
SUGGESTED ANSWERS
TnL SUPPORT MATERIALS
CHAPTER 1 : REDOX
Objective Questions
1 A 2 D 3 C 4 C 5 A 6 D 7 C
8 C 9 C 10 D 11 C 12 A 13 B 14 D
Structured Questions
1 (a) (i) Chemical energy to electrical energy 1

(ii) Electrical energy to chemical energy 1
(b) Allow the transfer of ion 1
(c) (i) P // Magnesium 1
(ii) Mg → Mg2+ + 2e 1
(iii) Mg(s) | Mg2+(aq) || Cu2+(aq) | Cu(s) 1
(iv) E0 = +0.34 – (-2.38) 1
= +2.72 V 1
(v) Reduction 1
(vi) Copper(II) ion // Copper(II) chloride 1
(d) (i) Anode : R 1
Cathode : S 1
(ii) Cu2+, Cl-, H+, OH- 1
(iii) Copper(II) ions // Cu2+ ion 1
(iv) E0 of Cu2+ ion is more positive than E0 of H+ ion 1
Cu2+ ion gains 2 electron to form copper atom 1
(v) Cu2+ + 2e → Cu 1
(vi) R : Greenish yellow gas bubbles released 1
S : Brown solid deposited 1
(vii) Blue colour become paler 1
Concentration of Cu2+ ion decreases 1
(e) 1
Magnesium ribbon Copper strip

Carbon electrode
karbon
Porous pot Copper(II)

Magnesium chloride
chloride solution
solution Diagram 6
(f) +2 to 0 1
CHAPTER 2 : CARBON COMPOUND
OBJECTIVE QUESTIONS
1 C 6 B
2 D 7 B
3 C 8 B
4 A 9 C
5 C 10 C
STRUCTURED QUESTIONS
1 a) 1. But-1-ene burns more sooty than butane 1

2. % C in but-1-ene = 4(12) x 100% = 85.71%
4(12)+ 1(8) 1
3. % C in butane = 6(12) x 100% = 82.76%
4(12)+ 10(1) 1
4. Percentage of carbon in but-1-ene is is higher than butane 1
b) 2C4H10 + 13O2 → 8CO2 + 10H2O 1+1
Number of mol of butane = 0.6 dm3 = 0.025 1
24 dm3 mol-1
From the equation : 2 mol C4H10 : 8 mol CO2 1
0.025 mol C4H10 : 0.1 mol CO2
Volume carbon dioxide = 0.1 mol x 24 dm3mol-1 1
= 2.4 dm3
c) (i) 1. Process I : Hydration 1
2. Temperature : 3000C 1
3. Pressure : 60 atm 1
4. Catalyst : Phosphoric 1
(ii)
Butan-1-ol Butan-2-ol
1+1
1+1
2-metihylpropan-1-ol 2-methylpropan-2-ol (Any 2 isomers)

d) 1
Ester
1
Butil etanoat
1
TOTAL 20
2 a) (i) Molecular formula shows the type and actual number of atoms of 1
each element in a molecule.
(ii) CnH2n+1OH 1
b) (i) Oxidation 1
(ii) Acidified potassium manganate(VII) solution 1
(iii) Butanoic acid 1
c) (i)
1+1
(ii) C4H5OH + 2[O] → C3H7COOH + H2O 1+1

9
Chapter 3 : Thermochemistry
OBJECTIVE QUESTIONS
1 C 2 D 3 D 4 C 5 B 6 C
7 B 8 C 9 C 10 C 11 C 12 B
13 B 14 D 15 D 16 B 17 B 18 B
STRUCTURE QUESTIONS
Question Answer Marks

1 (a) Heat of displacement is the heat change when one mole of a metal is 1
displaced from its solution by a more electropositive metal.
(b) (i) Use a polystyrene cup / plastic 1
(ii) Blue colour of the solution turns pale // 1

Brown deposit formed
(c) 1. No heat change 1
2. No reaction // Silver is less electropositive than copper. 1
(d) (i) 1. Heat change, Q

2. Change in temperature, θ.
3. Highest temperature with correct units.
100
Number of moles of solution = 0.5 x = 0.05 mol
1000
Q = 0.05 mol x 42 kJ mol-1 = 2.1 kJ = 2100 J

1
Mass of solution = 100 cm3 x 1 g cm3 = 100 g
Using Q = mcθ
Q 2100
θ= = = 5 0C 1
mc 100 𝑥 4.2
Highest temperature = (28 + 5) 0C = 33 0C

1
(ii) 1. Number of moles of copper(II) sulphate solution.

2. Mass of magnesium with correct units.
100 1
Number of moles of copperII) sulfat solution = 0.5 x = 0.05 mol
1000
1
Mass of magnesium = 0.05 x 24 g = 1.2 g
2 (a) To decrease heat loss// 1
Polystyrene is heat insulator //
Weak conductor of heat
(b) (i) Exothermic reaction 1
(ii) Is less /smaller // 1

(iii) Add the solution immediately/at once 1
Stir the mixture.
(c) (i) 25 1
Number of moles of Ag+ ions = 0.5 x 1000 = 0.0125 mol
(ii) Q = mcθ
= 50 x 4.2 x (31.5 – 29.0) 1
= 525 J // 0.525 kJ 1
(iii) 0.125 moles of Ag+ ions releases 0.525 kJ kJ heat

0.525 kJ 1
There for 1 mole of Ag+ ions releases = = 42 kJ mol-1
0.0125 mol
∆H = - 42 kJ mol-1 1
(d) heat is lost to the surroundings //Polysterene cups absorbs heat.. 1

3 (a) The heat of neutralisation is the heat change when one mole of water is 1
formed from the reaction between an acid and an alkali.
(b) (i) The mixture causes the container to become hot// temperature increase// 1
thermometer reading increases.
(ii) 1. Heat of neutralisation between sodium hydroxide solution and 1

hydrochloric acid is higher.
2. Hydrochloric acid is a strong acid /ionises completely in water 1
while ethanoic acid is a weak acid/ ionsies pertially in water.
3. Some of the heat released is used ionise ethanoic acid molecules
completely. 1
(c) Energy axis labeled with 2 differect energy level correctly drawn for 1
exothermic reaction.
Formulae of reactants and products correct 1
Label ∆H with correct value and negative sign correct.. 1
NaOH + HCl
Energy
∆ H = - 57.3 kJ mol-1
NaCl + H2O
(d) 1.Number of moles of water (H+ ions or OH- ions)
2. Heat released, Q
3. Temperature change with correct unit.
100 1
Number of moles of H+ / OH- ions= 1.0 x = 0.1
1000l
1
Q = 57.3 x 0.1 =5.73 kJ = 57300 J
1
5730
Temperature change = = 6.8 0C // 6.82 0C
200 𝑥 4.2

4 (a) A chemical reaction that releases heat to the surrounding. 1
(b) 1
Heay
(c) (i) The heat of combustion of propane is higher than methane.// or vice 1
versa.
(ii) 1. The number of carbon atoms per molekul of propane is more than 1
methane..
2. When the number of carbon atoms increases, the combustion of 1
propane will also poduce more carbon dioxide and and water.
3. There fore more heat is produced. 1
(d) Molar mass of propanol, C3H7OH = 60 g mol-1

60 g of C3H7OH burn produces 2016 kJ heat 1
There fore , 1 g of C3H7OH burn produces = 2016 kJ 1
60 g
= 33.6 kJ g-1 1
(e) Put a cold pack on the injured area. 1
This will absorbed heat from the injured area. 1
Blood vessels will constrict and will slow down the blood flow/ fluid 1
formation will be less on the injured area,
Nota: any endothemic reaction which will relieve the pain on the injured
area.
5 (a) (i) Use a metal container not a beaker. 1
Replace the wire gauze with a pipe clay trangle. 1
(ii) Heat released, Q = mcƟ

= 200 x 4.2 x 30 1
= 25200 J / 25.2 kJ (with units, J / kJ) 1
(iii) 1. Number of mole = 1.72 // 0.02 1

86
2. Heat of combustion, ∆H = 25.2 1
0.02
3. = -1260 kJ mol- 1 negative sign with units ( kJ mol- 1) 1
(d) 1. Fuel value of ethanol = 1376 = 29.9 kJ g-1 1

46
2. Fuel value of butan-1-ol = 2675 = 36.1 kJ g-1 1
74
Butan-1-ol is a better fuel because 1 g butan-1-ol releases 36.1 kJ of heat. 1
Answers Chapter 4 T n L Support Materials
Objective questions
1 C 2 A 3 B 4 C 5 A 6 D
7 A 8 B 9 C 10 B 11 A 12 C
Subjective questions
1 (a) Condensation polymeriation. (1)
(b)
(1+1)
(b) (i) Inert(1)
which enable to store hazardous chemicals (1)
(ii) Long lasting and hard to decompose naturally. (1)

It takes a very long time for the polymer disintegrate once disposed and can cause
pollution. (1)
2 (a) (i) Hydrogen ion (1)
(ii) Ethanoic acid (1)
(b) (i) Rubber particles collide with one another and breaks the protein membrane. (1)
Rubber polymers combine that cause latex to coagulate.(1)
(ii) Add ammonia solution into latex.(1)

Chapter 5 :CONSUMER AND INDUSTRIAL CHEMISTRY
OBJECTIVE QUESTIONS
1 A 2 B 3 B 4 B 5 B 6 C
7 C 8 A 9 d 10 D 11 C
STRUCTURED QUESTIONS

1 (a) (i) Fats are esters produced through the reaction between fatty acids and 1
glycerol
(ii) 2
Characteristics Fats Oil
Source Animal Vegetable
Physical state at Solid Liquid
room temperature
Melting point High Low
Fatty acid content High percentage High percentage of
of saturated fatty unsaturated fatty acids
acids
any two
(iii) 1. Supplying energy, 2

2. Providing body temperature insulation and
3. Helping with the absorption of important vitamins.
any two
(iv) 1. Excessive intake in our diet can contribute to heart related 1

problems,
2. Weight issues or obesity. 1
3. The risk for arteriosclerosis or hardening of the arteries if excessive 1
fat intakes are sourced from animals or saturated fats.

2 (a) (i) X : Soap Y : Detergent 2
(ii) Cleaning agent Y. 1

Does not form scum 1
(iii) Cleaning agent X 1
Made from natural resources 1
(iv) Hydrophilic part 1
(b) Biological enzyme- To remove protein stains, such as blood, milk 1
and sugar.
Whitening agent-To change dirt to colourless substance. 1
4
3 (a) (i) Type of food additive: Preservative 1
Function: Prevent or delay the growth of bacteria or fungi to make the 1
food last longer.
(ii) • allergies 1
• nerve disorder
• cancer
• asthma
• rashes
• hyperactivity in children
any one
(b) (i) Salt will draw out water from the cells of microorganisms and will 1
retard the growth of bacteria or fungi so that food can be kept longer 1
(ii) • Sugar . 1
• Vinegar
any one
(c) (i) Pectin/ Lecitin 1
Pectin- stabiliser 1
Lecitin -emulsifier
(d) Growth of bacteria or fungi 1

Oxidation of food 1

4 (a) (i) • Helps in the treatment of hair problems 1
• Thickens the hair
• Blacken the hair
(ii) Analgesic 1
(iii) Parcsetamol 1
(iv) Liver damage 1
(b) Mercury 2
Hydroquinone
Betamethasone valerate
Tretinoin
any two
(e) Mercury - Skin irritation and damage to kidney and nervous system if 2
absorbed into the bloodstream..
Hydroquinone - Hypersensitive skin and exposure to harmful UV rays
caused by reduced pigmentation.
Betamethasone valerate - Skin irritation and changes to skin pigmentation.
5
Tretinoin - Redness and peeling skin

5 (a) (i) 1. Diamond 1
2. Graphite 1
(ii) 1. Hard and strong 2

2. Transparent
3. Good conductor of heat and electricity
4. Elastic
5. Non-permeable
6. Low electrical resistance
any two
(iii) 0.1 nm 1
(iv) Electronic - conductor 2

Polymer and composite -polymer composite materials.
Membrane -Water filtration. Separating water from gaseous mixtures
Energy -Batteries that last longer, flexible and strong, supercapacitor
Censor- Graphene has high surface area.
Biomedical - Censors, tissue engineering, medicine delivery system
any two
(b) (i) 1. Ensuring more efficient waste management, 2

2. Reduction in greenhouse gases.
3. Emission and removal of cleaner wastewater.
any two
(ii) electrocoalgulation 1
(iii) copper / carbon 1

Understanding Redox Reactions

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Understanding Redox Reactions

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SUGGESTED ANSWERS

Laboratory Activity 1A (Page 7)

Chlorine water Potassium iodide

Activity 1E (Page 14)

Laboratory Activity 1B (Page 15)

Laboratory Activity 1C (Page 18)

Laboratory Activity 1D (Page 20)

Activity 1F (Page 26)

Self Assess 1.2 (Page 26)

Reducing agent : Ag, Cu, Zn, Mg

Experiment 1A (Page 29)

Activity 1G (Page 30)

Self Assess 1.3 (Page 30)

Apparatus set-up Observation Inference Conclusion

Bulb does not light up No free moving ion Glucose, C6H12O6 is

Bulb light up Presence of free Asid oksalik, C2H2O4

Bulb light up Presence of free Ammonia , NH3 is

Laboratory Activity 1E (Page 34)

Activity 1I (Page 37)

Laboratory Activity 1F (Page 38)

Experiment 1B (Page 41)

Hydrochloric acid, HCl 0.0001 mol dm−3

Activity 1J (Page 43)

(c) Product if concentration of electrolyte 0.0001 mol dm−3

(c) Product if concentration of electrolyte 0.0001 mol dm−3

Activity 1K (Page 44)

Laboratory Activity 1G (Page 45)

Nikel Iron ring

Nikel nitrate solution

Self Assess 1.4 (Page 47)

Impure nikel Pure nikel

Nikel chloride solution

(b) Anode : Impure nikel metal becomes thinner.

Self Assess 1.5 (Page 51)

Experiment 1D (Page 52)

1. To detect the present of Cu2+ ion and Fe2+ ion.

Activity 1M (Page 57)

Self Assess 1.6 (Page 57)

Achievement Test (Page 59)

Enrichment Corner (Page 60)

Activity 2A (page 66)

c) Organic compounds containing only hydrogen and carbon. Hydrocarbon

Laboratory activity 2A (page 69)

Self asses 2.1 (page 70)

1. The first six members of alcohol

The first six members carboxylic acid

Activity 2G (page 78)

Hydrocarbon Non Hydrocarbon

Alkane Alkene Alkyne Alcohol Carboxylic

CnH2n+2 CnH2n CnH2n-2 CnH2n+1O CnH2n+1 CnH2n+1 COO

ii) Substances that exist in the form of liquid : P,R,S,andT

c) - Boiling point S is higher than Q

C2H4 + H2 O H3PO4 C2H5OH

- Addition of acidified potasium manganate(VII)

C2H4(g) + H2O [O] C2H4(OH)2

2 a) - Hexene burns more soot than hexane

Laboratory activity 2C page 92

3 C2H5OH + 2[O] → CH3COOH + H2O

1 a) 2CH3OH + 3O2 → 2CO2 + 4H2O

Laboratory activity 2D page 96

3 b) 2CH3COOH + Na2CO3 → 2CH3COONa + H2O + CO2

c) 2CH3COOH + CuO → (CH3COO)2 Cu + H2O

(i) Molecular formula of alcohol component :CH3OH